
Class^Q /i JllL 
Book ^S^'^A^^. 



Cpipght]^". 



COKRIGHT DEPOSIT 



i 



Elements of Plane and Spherical Trigonometry, 
second edition, 178 pp., 8vo ...$1.00 

Table of Logarithms and Trigonometric Func- 
tions, to five places of decimals, xxxii + 76 
pp., 8vo : 0.75 

Elements of Trigonometry and five-place Tables, 
bound together 1.25 

A Short Course in Plane and Spherical Trigo- 
nometry, with four-place Tables, 116-[-xxviii 
pp., 8vo 1.00 

The Same, without Tables, 116 pp., 8vo 0.90 



Author and Publisher, 
EDWIN S. CRAWLEY, 

University of Pennsylvania, 

Philadelphia, Pa. 



A SHORT COURSE 



IN 



PLANE AND SPHERICAL 



TRIGONOMETRY 



BY 

EDWIN S. CRAWLEY, Ph.D., 

Thomas A. Scott Professor of Mathematics in the University of Pennsylvania 



E. S. CRAWLEY 
UNIVERSITY OF PENNSYLVANIA 
PHILADELPHIA 
1902 



THE LIBRAf?Y OF 
CONGRESS, 

T^'Jr) Copies Recsived 

OCT: '(0 ^902 

COPVRMJHT ENTRY 

CLASS ^XXa No. 

A^ D 1^ U- Lj- 

OOPY 3. 



Copyright, 1902, 
By Edwin S. Crawley. 



PRESS OF 

THE NEW ERA PRINTING COMPANY, 

LANCASTER, PA. 



a 






( 



PREFACE. 

This book has been prepared with the view of giving a 
very brief account of the fundamental principles of trigo- 
nometry, along the same general lines as those followed in 
the author's Elements of Plane and Spherical Trigonometry. 
It is in part an abridgment of the larger book ; but at the 
same time some new features are introduced with the aim of 
rendering more thorough the rapid assimilation, on the part 
of the student, of the ideas and methods which he meets in 
trigonometry for the first time. This is accomplished mainly 
by the frequent use of simple exercises, which are intended 
to bring clearly and forcibly to the student's attention the 
nature and uses of the trigonometric functions. 

E. S. C. 

University of Pennsylvania, July, 1902. 



CONTENTS. 



CHAP. PAGE 

I. — Introductory Topics 7 

II. — The Trigonometric Functions 14 

III.— Solution of Right Triangles 40 

IV. — General Formulae 49 

v.— Solution of Oblique Triangles 60 

VI. — Miscellaneous Problems 79 

VII.— Spherical Right Triangles 85 

VIII. — Formulae for Spherical Oblique Triangles 96 

IX. — Solution of Spherical Oblique Triangles. . ........ , . . . , 108 



A SHORT 
COURSE IN TRIGONOMETRY. 



CHAPTEE I. 

INTRODUCTORY TOPICS. 

1. Measurement of Angles. To measure a quantity it 
is necessary to employ a unit of the same kind as the quan- 
tity to be measured. Thus, to measure length we employ a 
unit length, as an inch, a foot, a yard, a mile ; to measure 
surface we employ a unit surface, as a square foot, an acre, a 
square mile. Similarly, to measure an angle we must employ 
a unit angle. Any angle whose size is definite may be em- 
ployed as the unit, for example, the right angle. A familiar 
example of this use of the right angle is in the determina- 
tion of the area of a spherical triangle, where the excess of 
the sum of the angles over two right angles, the right angle 
being the unit of measure for angles, equals the area of the 
triangle in terms of the area of a trirectangular triangle as 
unit of area. For ordinary purposes, however, it is found 
convenient to use a smaller unit than the right angle. Of 
such units there are three in practical use. 

I. The degree, which is one ninetieth (J^) of a right 
angle. 

II. The grade (pronounced grad), which is one hundredth 
(lio) of a right angle. 

7 



8 A SHORT COURSE IN TRIGONOMETRY. 

III. The radian, which is the angle subtended by an arc 
equal to the radius on any circle whose center coincides with 
the vertex of the angle. 

The degree (°) is of very ancient origin, descending to us 
from the Chaldeans. It is divided into sixty (60) parts 
called minutes ('), and the minute is divided into sixty (60) 
parts called seconds ("). When it becomes necessary to ex- 
press still smaller divisions of the angle, decimals of a second 
are used. Thus we write 36° 12' 42''.93. This system of 
angular units is called the sexagesimal system^ on account of 
the division of the degree and minute into sixty parts. 

The grade (^) is of French origin. It has some advan- 
tages, but its use has gained little currency outside of France. 
The grade is divided into one hundred (100) parts called 
minutes ('), and the minute into one hundred (100) parts 
called seconds (''). Smaller divisions of the angle are desig- 
nated, as in the sexagesimal system, by decimals of a second. 
Thus we write 57^ 3' 72'\3. This system is called the cen- 
tesimal system. The angle just written can be expressed also 
as 57.^03723, or, since the grade is one hundredth of a right 
angle, as .5703723 of a right angle. 

When it is necessary to distinguish between minutes and 
seconds of the two systems they are designated respectively 
as sexagesimal and centesimal minutes and seconds. 

Exercise. Show that 81" = 250". 

The radian is the unit of that system of angular measure 
which is called the circular system. It finds its principal ap- 
plication in the general discussions of the higher branches 
of mathematics. 

2. Relation between the Sexagesimal and Circular 
System. Geometry teaches that in the same circle or equal 
circles angles at the center are proportional to their inter- 
cepted arcs. Let P0§ be an angle of one radian, and BAC 



INTRODUCTORY TOPICS. d 

any other angle, which we will suppose contains radians. 
With and A as centers and equal radii describe arcs BS 
and DE, then 

BAC BE DE 
~ POq '~BS~ AD' 




D B 



because RS = AD by hypothesis. Then if length of arc 
DE = I and AD = r, we have 

6 = '-■ (1) 

Now let I = Si semi-circumference, then I = ;rr, where 
TT = 3.14159 •• •, and substituting this value of I in (1), we 
have d = TT. That is there are tt radians in an angle sub- 
tended by an arc of a semi-circumference. In other words 

180° = ;r radians. (2) 

Hence if x is the number of degrees in any angle, and d the 
number of radians in the same angle, that is if ic° = d radians, 
we have 

180° -;r' ^'^) 

from which x can be computed if ^ is known, or vice versa. 

Exercises. 1. What is the circular measure of (that is, the 
number of radians in) 36° ? 

X 

From (3), d=---—Tr, and since re — 36°, we have ^ = xVo^ 
180 

= ^TT radians. 



10 A SHOET COURSE IN TRIGONOMETRY. 

2. What is the number of degrees in an angle of t^tt radians ? 
From (3), a; =-180°, and since ^^To^j we have x^~ 

X 180° = 75°. 

3. What is the circular measure of 40° 20^? 

Eeduce both terms of the fraction on the left in equation (3) to 
minutes. 41° 20^ = 2480^, and 180° = 10800^, hence 

2480 31 

TT = ^^^ TT radians. 



10800 135 

If we substitute for n its value, 3.14159-.., we have = . 7214 
radians. 

4. What is the number of degrees in an angle of 2.75 radians ? 
Putting ^ = 2^75 in (3), we have 

. = H:^l><i5^ = *8^ = 157°.56 = 157° 34'. 



3. Directed Length. The requirements of trigonometry 
and higher branches of mathematics make it necessary to con- 
sider what is sometimes called the sense of a given length. 
By this is meant the direction in which the length is sup- 
posed to be measured along the line in which it lies. Thus 

on the line in Fig. 2 the length 

A B AB is the same as the length 

j7jQ 2. ^^ ^^ magnitude, but JSA is 

opposite to AB in sense. This 
difference in sense is indicated by a difference in sign, and 
hence AB = — BA. The reason for indicating a difference 
in sense by a difference in sign is based upon the following 
considerations. Suppose a point starts at A and moves to B, 
and then moves back from B to A. It has now traveled a 
distance equal to 2AB, but at the end of its journey its dis- 
tance from the starting point A is zero. Attending simply 
to this last consideration, we may say 

AB + BA = ; and hence AB = — BA. 



INTRODUCTORY TOPICS. 11 

Again, if three points A, B, C be taken in a straight line, 

we have 

AC= AB^BC, 

for the lengths here are + ^ 4- 

all of the same sign. We -p^^ ^ 

can say also 

AB = AC-\-CB, 

for 5(7= — CBy and hence this last is equivalent to 

AB = AC-Ba 

In the same way it appears that 

BA = BC-{- CA, BC=BA + AC, etc. 

4. Positive and Negative Angles. The principles ex- 
plained in the last paragraph, if applied to angles, lead to a 
distinction between positive and negative angles. The crank 
of a locomotive revolves in one direction when the engine is 
running forward, and in the opposite direction when running 
backward. These two rotations may be 
distinguished as positive and negative. 
So in Fig. 4, the angle A may be de- 
scribed by a line rotating from the posi- 
tion AB to the position A C, or vice versa. 
The first conception of the angle differs 
Fig. 4. " ^^ sense from the second, and we may 
write 
BAC= - CAB. 

In many cases with angles as with lengths we are con- 
cerned merely with the absolute magnitude of the quantities, 
and the sense or sign is not considered ; but, on the other 
hand, numerous instances occur where it must be taken into 
account. 




12 A SHORT COURSE IN TRIGONOMETRY. 

Note : It is usual to call that rotation positive which is oppo • 
site to the motion of the hands of a watch, and rotation in the 
reverse direction negative. This convention is observed through- 
out this book. 

5. Angles of Any Magnitude. An angle may be of 
any size indefinitely great, as the following example illus- 
trates. The minute hand of a watch in the course of twelve 
hours moves through 48 right angles, or 4320°. As it con- 
tinues to revolve, the angle through which it has moved 
subsequent to any given time goes on increasing indefinitely. 
The graphical representation of an angle greater than 360°, 
however, does not differ to the eye from that of an angle less 

^ than 360°. Thus in Fig. 5 ABC 

may represent one, five, nine, or 
in general 4^+1 right angles, 
NV\ where n is any whole number, posi- 

\\\ tive or negative. Let the student 

,'/ A verify this statement by assigning 
\\^ ^'-^^..'■''' // to n different values, and then start- 

^<:;.^_ ^^^:y ing at BA let him imagine a line 

^^'5 to rotate through as many right an- 

gles as the formula 4n -f 1 requires. 
He will find that it will always bring him to the position 
ABC. 

6. The Quadrants. For convenience of classification 
angles between 0° and 90° are said to be angles of the first 
quadrant, those between 90° and 180°, angles of the second 
quadrant, those between 180° and 270°, angles of the third 
quadrant, and those between 270° and 360°, angles of the 
fourth quadrant. Angles between 360° and 450° are angles 
of the first quadrant again, and so on. Similarly angles 
between 0° and — 90° are angles of the fourth quadrant, 
those between — 90° and — 180° are angles of the third 
quadrant, and so on. Eeferring to Fig. 6 it appears that if 



// 






1 



INTRODUCTORY TOPICS. 



13 



any angle has one of its sides placed in coincidence with 
OA, it is an angle of the first, second, third or fourth quad- 
rant, according as its other side b 
takes its position in the space 
marked I, II, III or lY. 



ir 



Examples. 



1. Express in radians 90°, 45°, 30°, A' 
60°, 135°, 110°, 75°, 63°. 

2. Keduce the following angles, 
given in radians, to the equivalent 
values in the sexagesimal system, Itt, 



III 



IV 



r^) fTT", T2^> i^) i^j T8" 



Fig. 6 



3. One radian = how many de- 
grees ? How many minutes ? How many seconds ? 

4. What is the value, in decimals of a radian, of 1°? Of 1^? 
OfV^? 

5. Reduce 23° 37^ to radians. 

6. Reduce 1.73 radians to degrees and minutes. 

7. On a railroad a portion of the track, 793 feet long, is curved 
to a radius of 1250 feet. What is the angle between the radii 
drawn to the extremities of the curve ? 

8. 150° is the same as what negative angle? 

9. 1000° represents how many complete revolutions, and how 
much over ? What quadrant is it in ? 

10. A line starts from the position OA, Fig. 6, and revolves in 
the positive direction through 579°; what is the least number of 
degrees through which it could have revolved to reach this posi- 
tion ? Would the revolution have been then in the positive or 
negative direction? 

11. In each of the following angles name the quadrant, and 
tell what is the least number of degrees, without respect to sign, 
which will express the inclinations between the sides of the 
angle : 

(a) 289°, (c) 651°, (e) — f tt. 



(b) - 172^ 



(d) 37r-f 60< 



(/) 27r-60°. 



CHAPTER II. 

THE TRIGONOMETRIC FUNCTIONS. 

7. Plane trig^onometry treats primarily of the relations 
between the six parts, the three angles and the three sides, 
of a plane triangle, so that when certain of these parts are 
known the others may be computed. This subject is treated 
in geometry also, but the methods there employed, being 
purely graphical, cannot be used to obtain accurate numerical 
results. Plane trigonometry investigates also the properties 
and relations of angular quantities in general, without refer- 
ence to triangles. 

The operations of trigonometry are carried on by the aid 
of certain quantities called trigonometric functions. To 
every angle belong six definite numbers which are called the 
trigonometric functions of that angle. The way in which 
these numbers are associated with the angle to which they 
belong will be explained in this chapter. Their names, and 
the abbreviations used in writing them, are sine (sin), COSine 
(cos), tangent * (tan), cotangent (cot, ctg, ctn, or cotan), se- 
cant * (sec), and cosecant (esc, or cosec). Thus 

sin 30° is read " sine of 30° '' or " sine 30° " 
cos 19° " " " cosine of 19° '' " " cosine 19° " 

* The words ''tangent" and ''secant" in trigonometry have an en- 
tirely different meaning from that in geometry, although there is an his- 
torical connection between the two. Trigonometry began as an outgrowth 
of geometry and the terms sine, cosine, tangent, etc., were the names given 
to certain lines associated with arcs of circles or with the angles at the 
center which these arcs subtended. Of these lines the ones designated as 
the tangent and secant were actually tangent and secant to the circle. 
The best modern usage however regards all the trigonometric functions 
strictly as numbers. 



THE TRIGONOMETRIC FUNCTIONS. 



15 



tan 72° is read " tangent of 72° '' or " tangent 72° " 
cot 40° " " " cotangent of 40° " " " cotangent 40°'' 



etc, 



etc, 



etc. 



8. The Sine and Cosine of any Angle. Tahe any point 
on one side of an angle and let fall a perpendicular from this 
point to the other side. 

The sine of the angle is the ratio of the length of the perpen- 
dicular to the distance of the top of the perpendicular from the 
vertex of the angle. 

The cosine of the angle is the ratio of the distance of the foot 
of the perpendicular from the vetiex to the distance of the top of 
the perpendicular from the vertex. 




Fig. 7. 

Let POQ be any angle d. Take any point on one side of 
the angle, as L on the side 0§, and let fall the perpendicu- 
lar LM to the side OP, produced if necessary. For brevity 
represent the length LM by a, the length OM by b, and the 
length OL by c. Then the sine and cosine of the angle 6 are 



sin d = 



a 



cos = 



Exercises. 1. Suppose, in Fig. 7, LM= 7 units and OL = 10 
units, then by the definition sin 6 = yV- Since OLM is a right 

triangle 0M= VoU 
= J^i/51 = .71414-... 



LM" = V 100 — 49 = t/51. Hence cos 



16 



A SHORT COURSE IN TRIGONOMETRY. 



2. Suppose 0M= 12 units and ML — 5 units, what are sin d 
and cos ^? 

3. Suppose in a right triangle (Fig. 8) 




a = 7 units and h = 
hj^potenuse c = W 
units. 



24 units, then the 



figure, 



24^ = 1/625 = 25 
Applying the definitions to the 



^^^^ = ^ = 25' 

. p h 24 
sm B = — = -—, 

c 25' 



cos A = 



24 
25* 



„ a 7 
cos ^ = - = - . 
c 25 



29 and a = 20, what are the sine 



4. In a right triangle let c 
and cosine of A and B ? 

5. Suppose the two legs of a right triangle are equal, then each 
acute angle = what ? What are the values of the sine and co- 
sine of one of the acute angles? 

It is to be observed that it makes no difference on which 
side of the angle the point L, Fig. 7, is taken, nor how far 
from the vertex. For the triangles OL^M^, OL^M^, OL^M^ 
(Fig. 9) are all similar right triangles, since they have one 

L.. 




Fig. 9. 



acute angle in common, and hence the ratios of their homolo- 
gous sides are equal ; that is, 



cos 6, 



L,M, L,M, 


h^^. 


OL^ ~ OL^ ~ 


OL, ~ 


031, OM^ 


OM, 


OL, - OX, ~ 


OL,- 



THE TRIGONOMETRIC FUNCTIONS. 17 

9. Properties of the Sine and Cosine. From Fig. 7 

a^ + 62 = g\ 
Dividing by c^ 

? + ?=!• 
Now 

Hence sin^ S + cos^ 6=1, 

from which sin = i/l — cos^ 0, > (4) 

and cos 6 = i/l — sin^ ^. 

That is, expressed in words : 

I. The sum of the squares of the sine and cosine of any 
angle equals unity. 

In the left hand part of Fig. 7, sin d and cos OLM 

both equal -, and cos d and sin 0L3I both equal - . But 
OLM= 90° - 6, hence 

sin e = cos (90° — 6), and cos 6 = sin (90° — 6). (5) 

Or, since 90° — ^ is called the complement of d : 

II. The sine of an angle is equal to the cosine of its com- 
plement, and vice versa. 

III. The values of sin and cos 6 are never greater than 
unity. 

For it is evident from Fig. 7 that a and b can never ex- 
ceed c, hence neither of the fractions - and — , which are re- 

' c c' 

spectively sin d and cos d, can exceed unity. 

*It is usual to write sin^ 6, rather than (sin 6)^ to indicate the square 
of sin Bj and similarly for other powers. It is read "sine square ^." 



18 A SHORT COURSE IN TRIGONOMETRY. 

Exercises. 1. When 6 = 90° what are the values of the frac- 
tions - and - ? Hence what are the sine and cosine of 90° ? 
c c 

2. Let it be given that sin 6 : cos d = 5-A, what are the values of 
sin 6 and cos^? 

From the given proportion we have cos 6 = ^ sin d, hence 

from (4) 

1 (\ 
sin2 e^ -~ sin^ 6 = 1. 
Zd 

41 5 4 4 

.•. -— sin^^ = 1, or sin ^ = , and cos^=psin^= -—==.. 

25 ' i/4l' 5 y^4i 

3. We can generalize exercise 2 by taking sin 6 : cos 6 = p : q. 

Then cos 6 =- sin 6. Hence, as before 
P 



(6) 



This problem is of great importance, and the results should be 
memorized. 

4. Given sin d : cos ^ = 8 : 13, find sin ft and cos ft. From the 
result of exercise 3 we have at once 

8 , 13 

sm ft = — ^=^, cos 6 = 



or 






sin^ 


d^ 


% sin^ ft = 
pi 


1, 


P' 


P' 


sin 


'6 = 


1, 


and 


sin^ = 


P ^ 




Vp' + q' 


from which 












COS^: 


q 


Vp^ ^- q^ i 



V 233 V 233 

Or the results may be obtained independently by the process used 
in exercises 2 and 3. 

5. Sin : cos e = 4 : 3. What are sin ft and cos ^? 

6. Sin ft : cos ft = l : 9. What are sin ft and cos ^? 

7. Sin ft : cos ft = V2 : Vl . What are sin ft and cos 0? 

10. The Tangent and Cotangent of any Angle. Con- 
struct a perpendicular as described in § 8. Then 



THE TRIGONOMETRIC FUNCTIONS. 



19 



The tangent of any angle is the ratio of the length of the 
perpendicular to the distance of the foot of the perpendicular 
from the vertex of the angle. 

The cotangent of any angle is the reciprocal * of the tan- 
gent. 

P P. 




Fig. 7 {Us). 



Thm 



tan = 1-, cot = —. 



Exercises. Take the five exercises on pages 15 and 16 and 
extend them to include the tan and cot of the angles, thus : 

7 



1. LM= 7 units, and 0M= i/51 units. Hence tan 6 = 



l/5l' 



and cot d = 



i/51 



2. 0M= 12 units, and ML = 5 units, what are tan 6 and cot ^? 

3. In the right triangle, Fig. 8, we have 

7 24 24 7 

tan A = -^, cot A = ^, tan B = ---, cot B = -—. 

24' 7 ' 7 ' 24 

4. In the right triangle in which c = 29, and a = 20, what are 
the tan and cot of A and B ? 

5. If the legs of a right triangle are equal, what are the tan and 
cot of one of the acute angles ? 



* The student is here reminded that the reciprocal of a quantity is 
unity divided by the quantity. Hence the reciprocal of a fraction is the 
fraction inverted. 



20 



A SHORT COURSE IN TRIGONOMETRY. 



11. Properties of the Tangent and Cotangent. 
I. We have directly from the definitions : 



cot^ = 



II. Since 



tan^' 
a a 



tan 6 = 



cot 6' 



tan^cot/?=l. (7) 



we have tan 6 = 
also cot 6 = 



c c 
sin 6 



cos d' 
cos 
sin 6' 



or tan 6 cos = sin ^, 



or cot ^ sin ^ = cos 6, 



(8) 
(9) 



III. Again in the left hand part of Fig. 7 tan ^ = y-, and 

cot OLM= p hence since OLM = 90° — ^, we have 

tan 6 = cot (90° - 6), cot 6 = tan (90° - 6). (10) 

That is : The tangent of an angle is equal to the cotangent 
of its complement, and vice versa. 

Exercises. 1. Construct an angle 
where tangent is f . 

On a straight line mark off any conve- 
nient length as AC. At C erect a per- 
pendicular, and mark off on it a length 
C~ CB=^AC. Then BAC is the angle, for 
BC 3 




Fig. 10. 



tan BAC = 



AC r 



2. Construct an angle whose cotangent is 3. 

Take ^C as before, and make CB = ^AC, then BACis the angle^ 

,or cot BAC^f^=^^=S. 

3. Construct the angles which have the following numbers, (1) 
for tangents, (2) for cotangents, viz. : |, |, 2, 5, |. 

12. The Secant and Cosecant of any Angle. 

l^he secant of any angle is the reciprocal of the cosine. 



THE TRIGONOMETEIC FUNCTIONS. 21 

The cosecant of any angle is the reciprocal of the sine. 
Hence we have, Fig. 7, 

c 

sec 6 = Tj CSC ^ = -. 
a 

Exercises. 1. In Fig. 7 if LM = 7 units, OL = 10 units and 

^— . 10 ' 10 

0M=^ V 51 units, sec 6 = --,=, esc 6 = -^. 

2. If OM— 12 units and ML = 6 units, what are sec ^ and 
esc 6. 

3. In the right triangle, Fig. 8, 

A 25 .25 ^25 ^ 25 

sec A = z^, CSC A = -—, sec B = -—, esc B = --r. 

24' 7 7 24 

4. In the right triangle in which c = 29 and a = 20, what are 
sec and esc of A and B ? 

5. If the legs of a right triangle are equal, what are the sec 
and CSC of one of the acute angles ? 

13. Properties of Secant and Cosecant. 

I. It follows directly from the definition that 

sec^= 7i, cos ^ = 7,, and cos ^ sec ^=1. HI) 

cos d sec ^ ^ ^ 

Also 

CSC 6 = - — 7i , sin ^ = ^ , and sin ^ esc ^ = 1. H 2") 

sin ^ CSC ^ ^ ^ 

II. The square of the tangent of any angle increased by 
unity equals the square of the secant. 

From Fig. 7 a^ + 6^ = c^, hence, dividing by P, 

^2 + 1 - p- 



But Y2 = tan^ dy and -^ = sec^ d, 

.'. tan^ ^ + 1 = sec2 d, and sec^ 6-1 = tan^ 0. (13) 



22 A SHORT COURSE IN TRIGONOMETRY. 

III. The square of the cotangent of any angle increased by 

unify equals the square of the cosecant. 

52 ^2 

Dividing a^ -f 6^ = c^ by a^, we have — 2 + 1 = — 3 . But 

-^ = cot^ ^ and -„ = csc^ 6, 
a^ a^ ^ 

.', cot' ^ + 1 = csc2 ^, and also csc^ |9 - 1 = cot^ (9. (14) 

IV. The secant of an angle equals the cosecant of its com- 
plement. 

In the left hand part of Fig. 7 we note that sec d = -j, and 
CSC OLM = |, where OLM = 90° - ^. Hence 

sec d = esc (90° — 6), and esc 6 = sec (90° — 0). (15) 

Note : The three pairs of formulae (5), (10), (15) may be com- 
bined in one statement, viz : Any function of an angle is equal to 
the co-function of the complement of the angle. 

Sin and cos, tan and cot, sec and esc are called co-functions of 
'each other respectively because originally the cosine was defined 
as the sine of the complemental angle or co-angle, and similarly 
with the others. 

14. The formulae which have been derived are here brought 
together under one view, in order that their similarities and 
differences may be emphasized. 



sin = cos (90° - 


-e), 


(5) 


tan ^ = cot (90°- 


-^), 


(10) 


sec 6 = CSC (90° - 


-6). 


(15) 


cot e = , tan d = ——, 

tan 6^ cot 6/' 


tan 6 cot ^ = 1, 


(7) 


sec 6 = , cos d = -, 

cos e^ sec e^ 


cos sec = 1, 


(11) 


^^^ ^ = ^rTh^ ^i^ ^ = ;:^r^' 


sin ^ CSC ^ = 1, 


(12) 



If 



THE TRIGONOMETRIC FUNCTIONS. 23 

sin 6 ' ^ cos d ,^^ ,^^ 

tan d = , cot e = -. — - . (8) (9) 

cos ^ ' sm 6/ V ^ V / 

sin^ d 4- cos^ G=:l, (4) 

sec2 ^ = 1 4- tan^ 6, (13) 

csc^ ^ = 1 + cot^ 6. (14) 



tan ^ =-^ = ^— , then sin B = -^ , cos 6 = ^ , (6) 

q cos 6 /p2 _|_ ^2 ^^2 _|_ ^2 

Exercises. 1. Given sin ^ — |, what are the values of the other 
functions of ^? 
From (4) 

cos d = Vl — sin^ = ^1 — 1 = ^1 z= i i/3 . 

From (8) 

1 





. r. sin ^ 2 11/- 
tan e = = = — = -1/3 




cos^ V3 ^^ ^ 




2 


From (7) 






cot e = - — - = /3 . 




tan d 


From (11) 






112 




sec (9 = =- = . 




cos ^ \ r~ /3 




2 


From (12) 






1 




CSC e = - — - = 2. 




sm Q 



2. Tan ^ = f , what are the values of the other fdnctions of ^? 

o- X /, sin ^ , sin ^ 2 „ 

Smce tan Q = , we have = -. Hence by (6) 

cose' cose 3 ^ ^ -^ 

2 3 

sin e = —=. , and cos 6 = — = . 
t/13 1/13 

Then, as in exercise 1, 

,^3 , v1^ . Vi3 

cot e — -, sec e = — — -, CSC 6 = -^r— . 

•<^ o 2 . 



24 A SHORT COURSE IN TRIGONOMETRY. 



3. Express sin d in terms of each of the other functions. For- 
mulae (4) and (12) give us first 

1 

csc^* 

1 



sin ^ = •/ 1 — cos^ 0, sin 6 = 



By (11) cos d = 



l/sec^ 



hence sin 6 = -v/l ^ ^ 

^ sec' 6 sec 

Then by (13) 



Vsed^ d — 1 = tan 6, and sec d =^ Vl 

. ^ tan^ 

hence sm d 



Vl + tan^ d 



1 



Finally, since esc 6 = Vl -\- cot^ 6, we have sin d = 

/1 + cot^^' 

4. Prove that ^ ^ „ 1 

sec d — tan d = -. 

sec 6 + tan 6 

Since from (13) 

sec^ 6 — tan^ 6=1 

we have 

(sec 6 — tan ^)(sec 6 + tan e) = l 

from which the statement above follows at once. 

5. Prove that 

sin 6 + tan 6 

sm 6 tan ^. 



cot ^ + CSC ^ 

Multiplying both sides of this equation by cot 6 -\- esc 6^ we have 

sin B -h tan e = ^\nd tan 6 cot (9 + sin ^ tan ^ esc 6, 

But by (7) and (12) tan 6 cot ^ = 1, and sin 6 esc Q = 1, hence the 
equation reduces to the identity 

sin 6 -h tan 6 = sin 6 -\- tan 6, 

which shows that the first equation is true. 

In each of the next six examples find from the given function 
the values of the other five. 



6. cos 6 = i/f . 


7. tan ^ = 3. 


8. sec 0=1. 


9. cot 6=^. 


10. CSC 6 = V2. 


11. tan^-- 
n 



THE TRIGONOMETRIC FUNCTIONS. 25 

12. Express tan d in terms exclusively of each of the other five 
functions. 

13. Do the same with sec 6. 

14. sin ^ = 2 cos d. Find the numerical values of each of the 
six functions of d. 

15. tan 6 = t/2 sin 6. Solve as in example 14. 

16. Prove 1 — sm 



1 + sin ^ 



- _ ^ sm 6 -f cos 6 . ^ ^ , ^ 

17. Prove —^ = sm sec + 1. 

cos^ 

, « ^ . 1 + tan d 

18. Prove sm 6 + cos Q = — . • 

sec^ 

19. Prove (sin 6 + cos oy + (sin 8 — cos df = 2. 

1 2 cos^ 

20. Prove tan ^ — cot (9 == .— . 

sm 6 cos 6 

21. Prove tan 6 -\- cot 6 = sec 6 esc 6. 

22. Prove ^-^^^^^ - tan 6, 

cot e + 1 

-- _, sec ^ + CSC ^ 1 + cot e 

23. Prove - = . — . 

sec ^ — CSC ^ 1 — cot ^ 

cot d cos 6 cot 6 — cos ^ 



24. Prove 



25. Prove tan ^ + cot 



cot 6 -f- cos d cot ^ cos 6 

sec^ ^ + csc^ 8 



sec ^ CSC ^ 



15. Versedsine and Coversedsine of any Angle. 

Besides the six trigonometric functions defined and discussed 

above there are two others, less frequently used, called the 

versedsine (vei^s) and coversedsine (covers). They are defined as 
follows : 

vers ^ = 1 — cos ^, (16) 

covers ^ = 1 — sin 8. (17) 



26 



A SHORT COURSE IN TRIGONOMETRY. 



16. Graphical Representation of the Trigonometric Func- 
tions. Let ACP = 0, then according to the definitions, 

B 




sin 6 = 



PM 
CP' 



AT 



sec 



CT 
~CA' 



Fig. 11. 



cos 



by (5) 

. p^_ SP CM 



by (10) 
by (15) 



cot 6 = tan PCB = 



CSC 



sec PCB = 



BR 
CB' 

CB 
CB' 



where it will be observed that the radius forms the denominator 
of each fraction. Hence if the radius be taken as the unit of 
length, 



sin = PM, 


tand = A T, 


sec = CT, 


cos = CM, 


cot e = BB, 


CSC 8 = CB. 



The student must be careful not to fall into the habit of think- 
ing of the trigonometric functions of an angle as lines. They 
are always pure numbers, but under the conditions established 
above they may be represented graphically by the lengths of the 
lines indicated. It is from this, the original, method of defining 
the trigonometric functions that their names are derived. 

The truth of certain of the formulae already established is 
clearly indicated by this diagram. Thus from 

MP' + CM' = CP\ we have sin^ 6 + cos^ ^ = 1, 

and from CT' — AT' = CA', we have sec^ — tan^ ^ = 1, etc. 

17. The Functions of 30°. Let ^^(7 be an angle of 30°. 
At any point B on AB drop the perpendicular BC to AC. 
Prolong BC to B' , making CB' equal to BC and join AB' . 



THE TRIGONOMETRIC FUNCTIONS. 



27 



Then ABB^ is an equilateral triangle, since angle BAB' = 
ABB' = 60^ Hence BC =^ ^ BB' = \ AB. Therefore 

. ^^^ BC ^AB 1 




That is 



sin 30° 



2* 



In exercise 1, p. 23, have been de- 
rived the values of the functions of the 
angle whose sine is |-, hence these re- 
sults are the functions of 30°, and we have ^therefore 



Fig. 12. 



sin 30° =2? cos 30°= 2^^ tan30° = ^T/3, 



cot30°=i/3, sec 30° = ;^ t/3 , esc 30° = 2. 

o 

Again, since 60° is the complement of 30°, we have from 
(5), (10) and (15) 

sin 60°= 2 1^^ cos 60° = 2, tan 60° = 1/3, 

1 — 2 — 

cot 60° = ^ t/3, sec 60° = 2, esc 60° = -^ t/3. 



18. Functions of 45°. In a right 
triangle whose acute angles are 45°, the 
legs are equal. Hence 

BC 
tan 45° = -tci— ^f ^^^ 

cot 45° = : T^7r= 1. 




tan 45° 



From (13) 



sec 45° = i/l + tan^ 45° = i/2, 



28 A SHORT COUESE IN TRIGONOMETEY. 

and from (14) 



CSC 45° = i/l + cot^ 45° = t/2. 
From (12) 

sin 45° = j^^ = -yr. = ^ \/\ 

CSC 45° |/^ 2 ' 

and from (11) 

cos 45° = ■ -r^^ =---... = - i/2. 

sec 45° -|/2 2 

From the fact that 45° is its own complement we could 
have anticipated that each function would have the same 
value as its co-function. 

Note : The thoughtful student will be inclined to speculate at 
this point on the question of how in general the values of the 
trigonometric functions of angles, given in tables of natural func- 
tions, are derived. An explanation of the means by which these 
numbers are computed, however, does not form a proper part of a 
work of this kind. It is sufficient to say that the sine and cosine 
of an angle can be expressed by convergent infinite series, in terms 
of ascending powers of the numerical measure of the angle, ex- 
pressed in radians, by means of which the values of these functions 
may be computed for any given angle. With the sine and cosine 
given, the values of the other functions are readily computed. 

19. Functions of 0° and 90°. It readily appears from 
the left-hand figure in Fig. 7 that if 0§ approaches OF the 
angle FOQ approaches the limit Oj and at the same time a 
approaches and h approaches c. Hence we have 



c 

sin 0° = - == 0, cos 0° = - = 1. 



c c 



Then 



o sill 0° ^ ^^ 1 1 

tan 0° = ^ = T = ^^ co^' ^ = ; no = a = ^^ 

cos 0° 1 ' tan 

sec 0° = -^^ = - = 1 CSC 0° = -. — -7^ = - = CD . 

cos 1 sm 0° 



THE TRIGONOMETRIC FUNCTIONS. 29 

SiDce 90° is the complement of 0°, we have also 

sin 90° = cos 0° = 1, cos 90° = sin 0° = 0, 
tan 90° = cot 0° = oo, cot 90° = tan 0° = 0, 
sec 90° = CSC 0° = CO, esc 90° = sec 0° = 1. 

A few words of explanation of the quantity oo (infinity) which 

occurs here may be useful. Every equation of the form =p 

requires that m =pn. If m remains constant and w diminishes, 
p increases. Thus 

I 10 10 

1000, etc. 



10 , 


10 ,^ 


10 ,^^ 


10 


;^ = 1, 


— = 10, 


— = 100, 




10 


1 ' 




.01 



Now if n becomes 0, we have — = what ? Evidently there is no 
finite number p, however large p may be, such that we can write 
— = Pj for this will require that p y<, = m, which is absurd, for 
every finite number multiplied by gives for the product. 
Hence the quotient - , is called infinity. From these considera- 
tions we can see that two of the qualities of oo are (1) great 
magnitude, transcending that of any finite number, and (2) the 
quality, by which it differs from any finite quantity, however large, 
that when multiplied by it does not vanish. The product of oo 
by is indeterminate, for^= oo,f= oo,|= oo, etc., no distinc- 
tion being possible in the quotient in any of these cases. 

Note : The teacher who wishes to introduce the solution of 
right triangles at the earliest practicable moment may pass 
here to Chapter III., returning after it is finished to complete 
this chapter. 

20. Change of Sign of the Functions. The definitions 
of the trigonometric functions given in §§ 8, 10, 12 are per- 
fectly general and apply to any angle whatsoever. So far, 
however, we have not considered the possibility that these 
functions may be sometimes positive and sometimes negative. 



30 



A SHORT COURSE IN TRIGONOMETRY. 



Take the four angles AOC^, AOC,, AOq, AOC„ all hav- 
ing OA as one side. Suppose them to be measured around 
from OA in the positive direction. Then AOC^ is an angle 

of the first quadrant, AOC-^ 

positive direc- ^n angle of the second 

quadrant, and so on. For 

brevity let AOC^=- d^, 



tionfor angles 




AOG,= 



6. 



A 00^ = 6^, the subscript 
designating the quadrant. 



Draw M^L^y 



MJL,, 



^sh> 



M^L^ perpendicular to AA', 
These perpendiculars are 
measured upward for an- 
gles in the first and sec- 
ond quadrants, and downward for those in the third and 
fourth quadrants. Hence in the one case their lengths are 
positive quantities and in the other negative. That is : 

The length of the perpendicular is positive for angles in the 
first and second quadrants, and negative for those in the third 
and fourth quadrants. 

Similarly the lengths OM^, OM^, OM^, OM^ measured 
from the vertex of the angle to the foot of the perpendicular 
are measured to the right for angles in the first and fourth 
quadrants, and to the left for angles in the second and third 
quadrants. Hence these are also sometimes positive and 
sometimes negative. That is : 

The length from the vertex of the angle to the foot of the per- 
pendicular is positive for angles in the first and fourth quad- 
rants j and negative for those in the second and third quadrants. 

The radial lengths OL^, OL^, etc., being measured in 
every direction from 0, according to the size of the angle, 
are for the present purpose without sign, and will be treated 
always as positive. 



THE TRIGONOMETRIC FUNCTIONS. 



31 



If the length of the perpendicular corresponding to any 
angle be represented by a, the length from the vertex to the 
foot of the perpendicular by h, and the length from the 
vertex to the top of the perpendicular by c, as in Fig. 7, it 
appears from ^ , 



sin 6 = 



and cos 6 = 



that sin ^ will have the same sign as a, and cos the same 
sign as b. That is : 

Sin is positive if 6 is in the first or second quadrant, and 
negative if 6 is in the third or fourth quadrant. 

Cos is positive if d is in the first or fourth quadrant, and 
negative if 6 is in the second or third quadrant. 

Since tan = t,, it follows that tan 6 is positive if sin d 

cos a 

and cos 6 have the same sign. Otherwise it is negative. 
Hence, 

Tan 6 is positive if is in the first or third quadrant, and 
negative if 6 is in the second or fourth quadrant. 

From (12), (11) and (7) it follows that the signs of esc 0, 
sec 0, and cot are always the same as those of sin 0, cos 0, 
and tan respectively. 

The following table gives these results in concise form. 



Signs of the functions of angles in different quadrants. 




First 
Quadrant. 


Second 
Quadrant. 


Third 
Quadrant. 


Fourth 
Quadrant. 


CSC^ 


+ 


+ 




— 


COS^ 

sec^ 


+ 


— 


— 


+ 


tan^ 
cot^ 


+ 


— 


+ 


— 



32 



A SHORT COURSE IN TRIGONOMETRY. 



Exercises. 1. Sin 6 is positive, in what quadrants will 6 be 
found? 

2. Tan 6 is negative, in what quadrants will 6 be found ? 

3. Sec 6 is positive, in what quadrants will 6 be found ? 

4. Cos 6 is negative, in what quadrants will 6 be found ? 

5. In what quadrant will 6 be found if 

(a) sin ^ is +, and cos 6 is — ? (c) cos 6 is — , and tan ^ is — ? 
(5) tan 8 is — , and sec ^ is + ? (d) cot ^ is +, and esc ^ is + ? 

21. Functions of 180° - d, 
180° + 0, and 360° - /? in terms 
of 0. Let AOC^ = e, and con- 
struct ^0G,= 180°-^. Then Ai M^ 
AfiC,=AOC^. Take OL^=OL^, 
then Jf2X2 = ifjii, and (}if2 = — OM^, Hence 



^ 




O Ml A 

Fig. 15. 



sin J.0C2 = -^^ 
cos AOC^ = Tyy^ 



-^^= sin AOC^, 






— cos AOC, 



That is 



.1 




sin (180° -&) = sin /?, 
cos (180° - /?)= -cos ^. 

L.^^^ Then by (8) and (9) 

tan (180° -d)= - tan Q, 
—^ cot (180° -0) = 



(18) 



cot 



}(19) 



Again, let ^0(7^ = d, and con- 
struct ^003 = 180° + 0. Then 
Fm. 16. j^^QQ^ _ ^^^^ T^kg ^^^ _ 

OX,, then MJj^ = - IT^X,, and OM^ = - Oif,. Hence 
sin AOC^ = ^^ = - ^^ = - sin AOC^, 



THE TRIGONOMETEIC FUNCTIONS. 



33 



COS J. 0(7, = -.^ 



— Y)T^ = "~ ^os ^OC^. 



That is sin (180° + /9) = - sin d, | 

cos (180° + (9) = - cos /?. J 

And, as above, tan (180° + ^) = tan ^, 
cot (180° + ^) = cot d 



(20) 



:} 



In fig. 17, AOC^^d, and 
^OC, = 360° - d. Then C,0^ 
= AOC^, Take OX^ = 0X„ then 
XjX^ is perpendicular to OA at ilf^, 
and ifjX^ = — MJj^. 
Hence 

sin^Oq=-^ = 
U-L. 




Fig. 17. 



cos AOC. 



-^= -sin^OCp 



That is 



sin (360° - /^) = _ sin d, 
cos (360° - /?) = cos d. 

And again, as above, 

tan (360° — /?)=— tan ^, 
cot (360° -/?)=_ cot /? 



:} 



(22) 



(23) 



22. Functions of 90° + d, 270° — d, 270° + /? in terms 
Oid. In formula (18) to (23) put /? = 90° - ^, and in 
the result write d for (p. Thus in (18) we have 

sin [180° - (90° - ^)] = sin (90° - (p), 

or sin (90° + ^) = cos f. 



34 



A SHORT COURSE IN a^RIGONOMETRY. 



As it makes no difference what letter is used to stand for 
the angle, this result may be written 



sin (90° -i- d) = cos 6, 
Similarly cos (90° 4- d) = — sin 6- 

And, from (19) 

tan (90° + /9) = - cot d, 
cot (90° + ^) = _ tan 6. 
From (20) 

sin (270° - d)= - cos d, 

cos (270° -d)=- sin 0. 

And from (21) 

tan (270° - 0) = cot d, 
cot (270° - ^) = tan 6. 

From (22) 

sin (270° -i-d)= - cos 6, 
cos (270° + (9) = sin 0. 

And from (23) 

tan (270° + <?)=- cot 0, 
cot (270° -{- 0)= - tan 0. 



(24) 
(25) 
(26) 
(27) 
(28) 
(29) 



The twenty-four formulae derived in §§ 21 and 22, can be 
summed up in two statements thus : 

function of (180° i 0), (360° - /?) = ± function 0, | 
function of (90° + 0), (270° ± 0) = ± co-function ^. J ^^^ 

In every case the proper sign must be prefixed to the result. 
(See § 20, table.) 

Note : In the formulae last derived d is supposed to be less than 
90°. The formulae are equally true for all values of 6, as will be 
shown later. They find their chief practical application in ex- 
pressing the functions of angles greater than 90° in terms of 
angles less than 90°. 



I 



Exercises. 1. Express sin 146° in terms of an angle less than 
90°. {a) If in (18) we write 180°— ^ = 146°, ^ = 34°, hence 



THE TEIGONOMETEIC PUNCTIONS. 



35 



sin 146° = sin 34°. (b) If in (24) we write 90° + ^ = 146°, d = 
56°, hence sin 146° = cos 34°. 

2. Express tan 321° in terms of an angle less than 90°. (a) If 
in (23) we write 360° — 6* = 321°, ^ = 39°, hence tan 321° = — tan 
39°. (6) If in (29) we write 270° + 6 = 321°, d = 51°, hence tan 
321° = — cot 51°. 

3. Express the following functions in terms of the same function 
of an angle less than 90°: sin 128°, sec 192°, cot 290°, tan 350°. 

4. Express the following functions in terms of a function of an 
angle less than 45°: esc 75°, tan 110°, cos 150°, sin 258°. 

5. Derive (24) from a figure. Let AOC^ = 6, and draw OC^ 
perpendicular to 0C„ then AOC^ = 90° + 0. Take OL^ = OL^, 
then M^L^ = OM^, and OM^ = — M^L^ 

Hence 

. .__, M„L„ 
sm AOC^ = -~~ 



OM. 



'2 OL, 

or sin (90° ^- 6) = cos 6. 

Similarly cos (90° + ^) = — sin d. 

6. Derive (26) from a figure. 

7. Derive (28) from a figure. 



cos AOC^. 




23. Functions of Negative Angles. In either figure let 
AOC^ = d, and AOC^= - 6. Take OL^ = OL,. 




L^G 



.woc. = S 




Fig. 19. 
ML. 



OL^ 



— sin AOC,, 



Ar.r. OM OM 

COS AOC^ = -^ = jjj- = COS AOC^. 



36 



A SHORT COURSE IN TRIGONOMETRY. 



(31) 



Hence sin (— ^) = — sin d, 

cos (— ^) = COS 6, 
tan (-/?)=_ tan 6, 
cot (- d) = — cot 6. 

24. Functions of 180°, 270°, 360°. 

I. Functions of 180°. As 

AOC approaches 180° ML ap- 

A preaches 0, and OM approaches 

OL, but being negative its value 




when J. 0(7 reaches 180° is — OL. 
Hence 





sin 180° = ^ = 0, 



tan 180° 



1 



cos 180° 



cot 180° 



OL 



OL 
- 1 
~0~' 






sec 180° = — - = - 1, CSC 180° = ^ = oo. 



- 1 







Note. — The student may be led to inquire why cot 180° is not 
written — oo. The cot of an angle a little less than 180° is very 
large and negative, while the cot of an angle a little more than 
180° is very large and positive. Just as it is changing from the 
second to the third quadrant, therefore, and has the value 180° it 
does not seem best to give the infinite value of the cot any sign. 
Some text-books give the value of cot 180° as ± <» , and similarly 
in other cases where like conditions exist. 

II. Functions of 270°. As ^0(7 approaches 270°, OM 
approaches 0, and ML approaches OX, but being negative 
its value when J. 0(7 reaches 270° is 

^MO 

OL 



— OL. Hence 

sin 270° = 



OL 



= -1, 



cos 270° = ^ = 0, 




Fig. 21. 



THE TRIGONOMETRIC FUNCTIONS. 



37 



tan 270° 



sec 270° 



- 1 
~0~ 

1 _ 

"" 



cot 270' 



CSC 270° 



1 = 0' 



1 



= - 1. 



III. Functions of 360°. The functions of 360° are the 
same as those of 0°. 

The following table which gives the functions of a few 
important angles will be found convenient for reference. 
The student will find it a profitable exercise to verify the 
values of the functions of angles greater than 90°, by 
means of (30) using the values of the functions of 30°, 45°, 
and 60°, given in §§ 17 and 18, as data. 



Angle. 


sin. 


COS. 


tan. 


cot. 


sec. 


CSC. 


0° 





1 





00 


1 


00 


30° 


i 


iV^ 


Jl/3 


1/3 


f/3 


2 


45° 


iT/2 


il/2 


1 


1 


/2 


/^ 


60° 


iT/3 


J 


l/3 


1/3 


2 


f/3 


90° 


1 





CO 





00 


1 


120° 


il/3 


-h 


-t/3 


-1/3 


-2 


f/3 


135° 


il/2 


-^t/2 


-1 


— 1 


-/2 


/2 


150° 


i 


-|/3 


-*/3 


-1/3 


-f/3 


2 


180° 





-1 





00 


-1 


00 


210° 


-i 


-iv-^ 


i/3 


/3 


-f/3 


-2 


225° 


-i/2 


-1/2 


1 


1 


-/2 


-/2 


240° 


-Jl/3 


—\ 


VI. 


i/3 


-2 


-f/3 


270° 


-1 





00 





oo 


— 1 


300° 


-iVs 


I 


-/3 


-1/3 


2 


-f/3 


315° 


-lV2 


i/2 


— 1 


— 1 


/2 


-/2 


330° 


-I 


j/3 


-il/3 


-/3 


|/3 


-2 


360° 





1 





00 


1 


c» 



Examples. 
1. In a right triangle the hypotenuse c = 41, and one side 
a = 9. Find all the functions of the angles A and B. 



38 A SHORT COURSE IN TRIGONOMETRY. 

2. COS 6 = — f , and sin 6 is negative. What is the quadrant 
of ^ ? Find the values of all the functions. 

3. tan = ^, and cos 6 is negative. Solve as in example 2. 

4. cos ^ = 3 sin 6. How many values less than 360° will d 
have ? What quadrants will they be in ? Determine the values 
of all the functions of each value of 6. 

5. tan = 2 cot 6. Solve as in example 4. 

6. tan 6 = — . Find all the other functions of B. 



t/1- 



m" 



7. tan e = x. Find all the other functions of 6. 

1 — m^ 

8. Express cos in terms of each of the other functions of 6, 

9. Find the value of each of the following expressions : 

l/ 3 cos 210° — t/ 2 sin 135° 



(a) 2 sin 90° — 3 cos 180°, {b) 



V 3 sec 330° + i/ 2 cos 225°' 
3 sin 270° (cos 360° + tan 45°) 



sec 120° (tan 315° + cot 135°) 

.r. ^. ,.. , , sin (90° + Q) tan (180° + B) 
lO.Simphfy (.) ^|r 80°-.)cot(270° + . )- 

(h\ tan (— 6) cos (180° + ^) cos (360° — &) tan (27 0° — Q) 

^ ^ cot (90°T ^) sin (— ey ^^^ sin (180° + B) cot (180° — B)' 

In each of the following examples as far as 21 find all the posi- 
tive values of B less than 360°. 

11. cos IB = J. 

Note: 2B= 60°, 300°, 420°, 660°; .-. 6* = 30°, 150°, 210°, 330°. 

12. sin IB = 1. 13. tan SB = — V^. 

14. sin^ 2B=l. 15. tan B = cot 23°. 

16. cos ^ = — sin 75°. 17. sin B = cos 150°. 

18. tan /9 -f cot (9 = 2. 19. sin^ B + cos B = ^. 

20. 5 sin^ B + cos^ 6* = 4. 21. 5 tan^ B + sec^ B = 7. 

22. sin (a + ^) = — i 1/3, cos (a — /3) = i- 1/2 . Find four pairs 
of positive values of a and j3 less than 360°. 



THE TRIGONOMETRIC FUNCTIONS. 39 

23. Trace the changes in magnitude and signs of the following 
expressions : 

(a) sin + cos 0, (b) sin 6 — cos 6, (c) 



sin d — cos 6 
Prove the following identities : 

f.. sin cos o • ^ 

24. H = 2 sm ^ cos 6. 

sec 6 CSC 6 

25. (tan 6 + cot d) sin 6^ cos ^ = sec^ d — tan^ 6, 

26. (tan 6 — cot 6) sin ^ cos ^ = 1 — 2 cos^ 6. 

nn n sin 6 COS 6 + cot 6 .^ 1 

27. cos = , . , ^ , ., , -. 28. tan d = , 

(sm^ ^ 4- 1) CSC ^ CSC (9 / 1 — sin^ 6 

tan 



29. sin e = + tan ^i/csc^ ^ — 1 — 1. 

sec 6 

2 1 — sin^i? 

30. cos = ^ -— . 

i/l + tan^^ cos^ 

31. tan d = a + cot^)(sec'^^-l) 

sin^ 6 + cos^ 6 + sin 6 sec 6' 

32. tan (45° =h ^) = cot (45° =f 6). 

33. cos (45° -^0) = — cos (135° — 6). 

34. sin (135° — 6) = — sin (225° + ^). 

35. tan (135° ^ e) = — tan (45° - 6). 

36. ^1^^' = CSC ^+2 cot ^. 

sin 6 

„_ 1 — tan^ f; cos ^ — sin ^ .^ • o ^ , o ^ 

37. ^ , ^ = . 38. sm^ d + vers^ 6 = 2 vers d. 

1 + tan 6 cos ^ 

39. sin* e + cos* ^ = 1 — 2 sin^ (9+2 sin* ^. 

40. sin*^ — cos*6' = sin^^ — cos^^ = 1 — 2cos2^ = 2sin2(9— 1. 

41. sec* 6 + tan* ^ = 1 + 2 tan^ 6 -{- 2 tan* 6. 

42. sec*^— tan*^ = sec'^ + tan'e=: 1 + 2tan26' = 2sec2^ — 1. 
2 sin^ ^ — 1 



43. tan* 6 — 1 = sec 



1 — sin^ 6 



. . tan a + tan B 

44. — ^ — ^ = tan a tan /?. 

cot a + cot /? ^ 

. - cot a + tan B 

^5- ^ r-^ = cot a tan /?. 

tan a + cot /3 

3 



CHAPTER III. 

SOLUTION OF RIGHT TRIANGLES. 

25. To Solve a Triangle means to find the values of the 
unknown parts from the given values of the known parts. 
Any triangle can be solved provided three parts, one of 
which is a side, are given. It follows that a right triangle 
can be solved if either a side and an acute angle, or two sides 
are known. The given quantities are called the data. 

In solving triangles the following principles should be 
observed. 

I. So far as it may be possible compute each unknown part 
directly from the data. 

II. It is desirable if possible to compute the different parts 
by more than one method. 

III. At the conclusion of the work use as a check some 
formula which expresses a relation between as many as pos- 
sible of the parts that have been found, and see whether 
the results obtained satisfy this relation. 

IV. If the results do not satisfy the check formula, go 
over the work again carefully and persistently until the error 
is discovered. 

Note : The student must bear in mind that the numbers used, 
and the results obtained, in arithmetical computations such as are 
performed in trigonometry, are always subject to a small residual 
error, due to the fact that the values of the trigonometric func- 
tions of angles, and logarithms, are incommensurable numbers ; 
that is they cannot be expressed exactly in whole numbers and 
decimals, except in a very few cases, such as sin 30°, tan 45°, etc. 
Thus when we find in a four-place table of natural sines that 
sin 25° = .4226 it means that the true value of this quantity lies 
somewhere between .42255 and .42265. The value given in the 

40 



SOLUTION OF RIGHT TRIANGLES. 



41 



table differs therefore from the true value of sin 25° by not more 
than .00005. In other words it is correct to within ^offoo of a 
unit. With a larger number of decimal places the degree of ac- 
curacy is correspondingly increased, so that results may be ob- 
tained to any practical degree of precision that may be desired. 



26. Solution of Right Triangles. 

Let ABC be any right triangle, and 
let a, hy c stand for the lengths of 
the sides opposite the angles A, B, C, 
respectively. Then from definitions of 
the trigonometric functions, we have 




sin A 



= cos B, 



cos ^ = - = sin ^, 



(32) 



(33) 



tan ^ = ^ = cot ^, 



(34) 



cot A 



= tan B, 



(35) 



Besides these we have a^ -\- b^ = c^, from which we derive 



or a= V{c - 6)(c + &). (36) 

Similarly h = V{c — a){c + a). (37) 

To find any unknown part of a right triangle we have 
only to select from these formulae one in which that part 
and two known parts occur, substitute the known values, and 
solve the equations. In the exercises which follow some are 
worked without and some with the aid of logarithms. 

Exercises. 1. Given ^ = 30° and b = 75 feet. Find a and c. 
From (34), a = b tan A, . •. a = 75 X 



From (33), c 



75 150 

~ Vs 



25 T/ 3 feet. 
= 50 i/3 feet. 



cos A' 1 i/3 

If the results are to be expressed decimally, we have, since 
1/3-1.732, a = 43.3, c = 86.6. 



42 A SHORT COURSE IN TRIGONOMETRY. 



For a check formula use h = V {c — a){G + a). 



V{c -a){c^a) = V25 t/3 X 75 1/3 = V75' = 75, 
which proves the correctness of the work. 

2. Given A = 60° and c = 24 feet. Find a and b. 
From (32), a = c sin ^, . •. a = 24 X 1 1/3 = 12 i/3 feet. 
From (33), h = c cos ^, . •. 6 = 24 X J = 12 feet. 
Using the same check formula as in Ex. 1, we have 



h = y{24: — 12 i/3)(24 + 12 t/3) = i/576 — 432 = VlU = 12. 

3. Give c = 53, and B = S8° 13\ Find a and b. The formulae 
are a = c cos B, b = c sin ^B. 

cos^= .7857 sin^= .6187 

c= 53 c= 53 

2.3571 lT856l 

39.285 30.935 



a = 41.6421 5 = 32.7911 

These values for a and b, however, are not reliable to four 
places of decimals. For, as has been explained, the values of 
sin B and cos^B may be in error to the amount .00005, and hence 
the error of a and b may be as much as .00005 X 53 = .00265. 
Hence not more than two decimal places in a and b are reliable. 
We write therefore a = 41.64, b = 32.79. 

For a check formula use a = V{c — 6)(c + b) = t/20. 21 X 85.79 
= i/r733.8i59 = 41.64. 

If logarithms are used in this example the work is as follows : 

log cos J5 = 9.8952 log sin 5 = 9.7915 log (c — 6) 1.3056 

logc = 1.7243 log c = 1.7243 log (c + b) L9334 

log a = 1:6195 log b = 175158 log a' = 3. 2390 

a = 41.64 ^) = 32.79 log a =1.6195 

4. Given a = 1.944, c = 3.007. Find A and b. 



a 



sin A= , b = c cos A, and b = V{c — a){c + a), check. 

loga= .2887 log c = .4782 log (c — a) = .0265 

log c = .4782 log cos^ = 9^825 log {c^a) = .6947 

log sin ^ = 9,8105 log 6 =^607 log 5^ = . 7212 

^ = 40° 16^ 6 = 2.295 log 6 =.3606 



SOLUTION OF RIGHT TRIANGLES. 43 

The difference in the values of log b arises from the small resid- 
ual errors in the logarithms which has been mentioned. The 
second value of log b gives b = 2.294, the true value lying prob- 
ably between the two. 

5. Given a = 639.4, b = 417.9. Find A and c. 



a a 



tan A = :r, G = -. — , , and a = Vic — 6 ) (c + b). check. 

log a = 2. 8058 log a = 2. 8058 log c — 5 = 2. 5390 

log 6 = 2.6211 log sin J. = 9.9228 log c + 6 = 3^0725 

logtan^ = 0.1847 log c = 2.8830 log ^2 = 5.6115 

^ = 56° 50^, c = 763.8, log a = 2.8058. 

Examples. 
Solve the following triangles without logarithms. 

1. A = 40° 23^, c = 16. 5. a = 5, b = 12. 

2. 5 = 28° 37^, b = 8. 6.5 = 59°, c = 5. 

3. ^ = 55° 16^, 6 = 7. 7.6 = 3.2, c = 8.7. 

4. a =8, c = ll. 8. a = 7.3, 6 = 2.4. 
Solve the following triangles with the aid of logarithms. 

9. ^ = 68° 13^, 6 = 63.27. 13. 6 = 62.49, c = 94.78. 

10. 5 = 51° 48^, a = 9.624. 14. a = .9362, c = 1.5621. 

11. ^=43° 29^, c = 362.9. 15. 5 = 12° 51^, 6 = 7.963. 

12. a = 3.675, 6=2.493. 16. ^ = 73° 42^ a = 1276. 

17. In an isosceles triangle the angle at the vertex is 56° 38^ 
and the base is 173 feet. Find the length of the side and the 
altitude. 

18. In an isosceles triangle one of the base angles is 74°, and 
the side is 9 feet. Find the base and altitude. 

19. In an isosceles triangle the base is 13.92 feet and the base 
angles are 69° 38^. Find the side and altitude. 

20. In an isosceles triangle the angle at the vertex is 54°, and 
the base is 76.2 feet. Find the length of the perpendicular from 
one end of the base to the opposite side. 



44 A SHORT COURSE IN TRIGONOMETRY. 

27. Special Solution when a Side and the Hypotenuse 
are Nearly Equal. Suppose h and c are given and are 
nearly equal, then A is very small and B is nearly 90°. 
Hence formula (33) which gives 

cos ^ = - = sin 5, 

will not determine either A or B with precision, because the 
cosine changes very slowly when the angle is near 0°, and 
the same is true of the sine when the angle is near 90°. 
To avoid this difficulty use (36), and then (32). 
Example : b = 479.1, c = 480.5. Find A and a. 

c-f6 = 959.6 log (c+ ft) = 2.9821 log a =1.5641 

c — b= 1.4 log(c — i)=_^]461 log c = 2.6817 

log ^2 ^ 3, 1282 log sin A = 8.8824 

a = 3. 665 log a = 1. 5641 ^ = 4° 22^ 

If the attempt is made to compute ^ by , , r, nr^r^A 



(33), the result is shown to the right, from 
which it appears that we can only say that 



log c = 2.6817 



A is somewhere between 4° 21^ and 4° 30^ ^^^ cos A — 9.9987 

Examples. 
Solve the following triangles. 

1. 5 = 60.74, c = 61.18. 3. & = 1.621, c = 1.635. 

2. a =103.7, c= 104.2, 4. a = 99.6, c = 100.0. 

28. Applications. 1. In a right triangle ,B = 63° 9% and the 
perpendicular from C to the hypotenuse in 161.2. Solve the tri- 
angle. 

Let p represent the length of the perpendicular. Then 

a 

P I P ^ 

Hence a = - — ^, o = -^ ^ j3 

sm B cos Br ^y^ ^ 



sin B= ^, and sin ^ = ^ = cos B. JC 



and from (32) c = = . 

^ ^ cos B 




For the check use b = V {c — a){c -{- a)- 




SOLUTION OF RIGHT TRIANGLES. 45 

logp = 2.2073 logi9 = 2.2073 log a = 2.2569 

log sin 5 = 9.9504 log cos 5 = 9.6548 log cos jB = 9.6548 

log a = 2.2569 log h = 2.5525 log c = 2.6021 

a = 180.7, c — a =219.3, log =2.3410 
6 = 356.8, c+a = 580.7, log = 2.7640 
c= 400.0, log 5 =2.5525. log 6^ = 5.1050 

Note : The angle of elevation of an object above the horizontal 
plane in which an observer is placed, or the angle of depression of 
.Q an object below that plane, is the angle 
which the line joining the object and the 
observer's eye makes with the horizontal 
^"^^^ ^:~7— :: ^plane. Thus, if E is the position of the 

observer's eye, is the object and EH a 
horizontal line in the same vertical plane 
Fig 24 "' with EO, the angle of elevation or de- 

pression is HEO. 

2. From the top of a lighthouse 110 feet above the level of the 
sea the angle of depression of a 

ship is 4° 14^. Determine the dis- ~=^=;;^~-----^^ 
tance of the ship from the foot of „ g 

the lighthouse. Fig. 25. 

Let LH be the lighthouse, and S the ship. 

Then tan Ja:SL=^, or HS = LH cot ESL = 110 cot 4:° U\ 
Ho 

log 110 = 2.0414 
log cot 4° 14^ = 1.1306 

log HS = 3.1720, .'.HS= 1486 feet. 

3. From the top of a building 50 feet high the angle of eleva- 
tion of the top of a flagstaff is 21° 38^, and the angle of depres- 
sion of its foot is 26° 34^ How high is the staff and how far is 
it from the building. 

Let AB be the building, and FS the flagstaff, then AB = 50 
feet, HBS = 21 ° 38^, HBF = 26° 34^ AF and FS are required. 

AF=AB cot AFB = AB cot HBF= 50 cot 26° 34^, 
HS = BHtaii HBS = AF tan HBS =■ AF tan 21° 38^, 
FS = FH + HS = 50 feet ^ HS. 



46 A SHORT COURSE IN TRIGONOMETRY. 

log^5 = 1.6990 log ylF= 2.0000 FH=50 

log cot HBF = 0.3010 log tan HBS = 9. 5984 HS = 39. 66 

log ^i^=: 2.0000 log HS = 1.5984: FS = S9.66 
^i^= 100 feet, 5^9=39.66. 

To check the work draw SP perpendicular to BF. Then 
SP = FS sin PFS, 

. _,p P„ . p-^_, BHsmPBS 

and SP = BS sm PBS = ^^^ , 

cos HBS ' 

where PBS = HBS + HBF = 48° 12^, 

and PFS = 90° — FBH = 63° 26^ 

If the two values of SP agree the work 

is right. The details of the logarithmic 

work are omitted. Fig. 26. 

Examples. 

1. One angle of a rhombus is 70°, and the shorter diagonal is 
13.8 feet. Find the lengths of the sides and of the other diagonal. 

2. The middle points of the sides of a rectangle form the ver- 
tices of a rhombus. Determine the angles of the latter if the. sides 
of the former are 2.789 and 4.163 respectively. 

3. Two tangents to a circle of radius 1.23 inches intersect 4.38 
inches from the center. Determine the angle between them. 

4. In a circle of radius 1, given the angle at the center 6. to de- 
termine the chord subtending the angle. 

(1) ^ = 43°2^ (2) ^ = 75°10^ (3) ^=125°18^ 

5. In a circle, given the radius r, and the angle at the center 
6, to determine the chord. 

(1) ^ = 27°48^ r = 4.176. (2) ^--168° 6^, r= 14.47. 

6. In a circle of radius 1, given the length of a chord s, to de- 
termine the angle at the center which it subtends. 

(1) 8 = 0.7492. (2) s = 1.7658. 

7. In a circle, given the radius r = 7.651, and a chord s = 8.446 
to determine the angle at the center subtended by the chord. 

8. Eegular pentagons are inscribed in and circumscribed about 
a circle of radius 100 feet. Determine the lengths of their sides, 
the apothem of the inscribed pentagon, and the radius of the cir- 
cumscribed pentagon. 



SOLUTION OF RIGHT TRIANGLES. 47 

9. Determine the same things for regular decagons inscribed 
in and circumscribed about the same circle. 

10. How great is the angle at the center of a circle, if the 
chord subtending the angle is two thirds of the radius? 

11. Determine the area of the segment of a circle, having given 
the radius r = 11,28 and the chord s = 6.97. (Note that there 
will be two solutions.) 

12. From a point in the circumference of a circle of radius 
r=6, two chords are drawn, whose lengths are respectively 
a =2.860 and 6 = 7.098. Determine the angle between the 
chords. (Note that the figure can be constructed in two 
ways.) 

13. The area of a regular heptagon is jP= 43.25. Determine 
the length of its side. 

14. The side of a regular dodecagon is 11.2 feet. Determine 
the radii of the inscribed and circumscribed circles. 

15. A regular enneagon whose area is 289.2 square meters is 
inscribed in a circle. What is the radius of the circle ? 

16. The area of a parallelogram is 5236 square feet, one side is 
396 feet and one angle 58° 19^. Determine the adjacent side. 

17. Determine the angle between the outer common tangents 
to two circles, having given the radii of the circles, B = 6.130, 
r = 2.014, and the distance between the centers = 8.49. 

18. In what ratio will the area of an isosceles right triangle be 
divided by the bisector of one of the acute angles ? 

19. A right triangle is divided into two equal parts by a 
straight line drawn from one of the extremities of the hypote- 
nuse. If the angle from whose vertex this line is drawn is 60°, 
in what ratio will it be divided ? 

20. In an isosceles trapezoid the area and parallel sides are 
respectively 3.477, 5, and 3. Determine the angles. 

21. The non-parallel sides of a trapezoid are respectively 3 51 
and 7.04 meters, and if produced they will intersect at right 
angles. Determine the angles of the trapezoid. 

22. Determine the angle of intersection of two circles whose 
radii are i^ = 527.4, r = 474.3, and whose common chord is 
a = 422. 

23. In a right triangle c — a = a — b. Find the angles. 



48 A SHORT COURSE IN TRIGONOMETRY. 

24. Upon the top of a cliff Ktands a j)ost 5 feet high. At a 
horizontal distanecj of 400 f(!(!t from the base of the cliff the top 
of the post haw an angle of elevation of 61° 2^. Determine the 
height of the clilf above the observer's eye. 

25. From the top of a lighthouse 324 feet above the level of 
the sea two ships appear in line. Their angles of dejjression are 
3° 5(/ and 2° 28^. Determine the distan(;e between them. 

26. UiJon the top of a tower 125 feet high stands a pole which, 
at a horizontal distance of 200 feet from the foot of the tower, 
subtcinds an angle of 3°. How long is the i)ole? 

27. A balloon, whose diameter is 2 meters, ascends vertically 
from a j)oint A. After it has risen a certain distance an observer 
at a horizontal distance of 800 meters from A observes that the 
balloon subtends an angle of 0° 4^. How high is it when the ob- 
servation is taken ? 

28. At A the angle of elevation of a cloud in the H.W. is 
43° IW. At B, 2526 meters due south from A, the cloud appears 
in the N.W. What is its angle of elevation at B, and what is its 
distance from the earth ? 

29. From the top of a lighthouse 224 feet above the level of 
the sea two ships are seen. Onci bears N. 72° E., and its angle 
of depressicm is 1° 19^, The other bears S. 18° E., and its angle 
of d(ipression is 2° 57^. Determine the distance between the ships. 

30. The length of a degree of longitude on the equator is ap- 
proximatcily 69.1 miles. Determine the length of a degree of 
longitude at Philadelphia, 40° N. latitude. 

31. Determine the area of an isosceles triangle in terms of the 
angle at the vertex (A), and the perpendicular from one end of 
the base to the opposite side (h). 

32. A regular pyramid has for its base a square whose side is 
458.2 feet. The angle between any one of its faces and the plane 
of the base is 59° 36^. Find its altitude, the angle its edges make 
with the i)lane of the base, its slant height, and the length of its 
lateral edge. 

33. The perpendicular distances of two points A and B from a 
straight line XY are 51 and 38, and the line AB produced makes 
with XY an angle of 23°. Find the radii of the two circles 
which pass through A and B and are tangent to XY. 



CHAPTER IV. 

GENERAL FORMULiE. 

29. The Sine and Cosine of the Sum and Difference of 
Two Angles. The formulae for sin {x rb y) and cos (x ± ?/), 
where .1- and y are any angles, are of such importance in trig- 
onometry that they are usually called 
"the fundamental formulae.^' The 
student should make a point of mas- 
tering their derivations completely. 

Ijet A OB = x, and BOC=y. Then 
AOC= X -\- y. At I), any point in 
OB draw DE perpendicular to OB. 
Draw EF and DG perpendicular to 
OA, and DH parallel to OA. Angle TIED = A OB 
We have then 

. , ^ FE GD^IIE (J I) HE 

sm [x + y) 

and 

_ or; HI) 
~ OE ~ OE ' 




X. 



0E~ 
X OF 



Now 



OE 



Oh 



a J) = OD sin X, HE = EJ) cos HEI) = ED cos x, 
00= 01) COS X, HD = EJ) sin HEB = ED sin x. 



Hence 

and 

But 



sin (.f + y) = 

cos(.T+2/) = 
OD 



OD 



ED 



^^,sma;+^^.cos.T, 



OD 



ED 



^j^.cosx-^j^^.mx. 



OE 



= cos y, 



and 
49 



ED , 

0E = ^y^ 



60 



A SHORT COURSE IN TRIGONOMETRY. 



and 




sin (oc + p) =: sin x cos y + cos x sin y^ (38) 

cos (x -{- y) = cos xcosy — sin x sin y. (39) 

B Taking AOB = x, and BOC=y, 

and constructing the figure as before, 
C with J)^ perpendicular, to OB, etc. 
we have AOC = x — y. Hence 

^ GE FD HD 



. . oa 

cos(x-y) = -^ 



OE OE OE 
OF + EG OF HE 



OE 



~ OE'^ OE' 



Then, since 

FD = OB sin x, HD = BE cos HBE = BE cos x, 
OF = OB cos X, HE = D^ sin HBE = D^ sin x. 



we have 



, OB . BE 

sin (^ - 2/) = -^ sm a; - -^ cos a;, 



But 

Hence 



cos (x — y) — 

OB 

= cos y, 



OB 



BE 



^^coso^ + ^^smo.. 



and 



BE 



sm 2/. 



(40) 
(41) 



OE ^' OE 

sin (x — y) = sin ir cos ^ — cos a? sin y, 
cos (x — y) ^ cos a? cos i/ + sin a? sin i/. 
These four formulae may be written thus : 

sin (xzhy) = sin x cos y d= cos x sin 3/, 

cos (x ^y) =: cos a; cos 2/ =f sin ^ sin y, 

where the upper signs on the two sides go together and also 
the lower signs. 

In the demonstration just given the angles x and y and 
their sum and difference were taken for simplicity as positive 



GENERAL FORMULA. 



51 



angles less than 90°. The formulae are perfectly general, 
however, and may be derived equally well if x and y have 
any magnitude, due consideration being given to the direction 
of the different lengths used in the demonstration. 

Let AOB = X, and BOC= y. 

From any point, (7, on OC draw CD perpendicular to OB 
(produced if necessary). Draw CF and DK perpendicular 
to OA, and HDG parallel to OA, Then 

COS {x + y)= -^. 

Now, taking into account the direc- 
tions of the lengths, 

0F= 0K-\- KF= 0K+ DG. 

But OK =0D cos X, 

and I)G= JDC cos HDC' Fig. 29. 

= DCcos (90° -j-x)= - DC sin x. 

0F= OD cos X - DC sm x , 




Hence 
and 

but 



OF OD 



cos {x + y) = -^= -^ cos 



DC . 

^^sm Xj 



OD 

-^ = cos2/. 



and 



DC 



g^=smy 

cos (x -]- y'J == cos X cos y — sin x sin y. 

Since y has any value, we may write — y for y, and the for- 
mula becomes 

cos {x — y) ^ cos X cos y + sin x sin y, by (31) 
Again 'writing 90° — x for x, we have, after reduction, 



* HDC measured in the positive direction of rotation = HDB + BDG 

■= 90° + X. 



52 



A SHORT COURSE IN TRIGONOMETRY. 



sin (^x — y) = sin x cos y — cos x sin y, 
sin (^x -{- y) = sin x cos 3/ + cos x sin 7/. 

Exercises. 1. Deduce the value of sin 15°. 
Since the functions of 45° and 30° are known, this can be done 
by using (40). Thus, putting x = 45°, y = SO°, 

sin 15° = sin (45° — 30°) = sin 45° cos 30° — cos 45° sin 30°, 

= i l/2 . i 1/3 - i t/2 ■ J, 

= i l/6 - i 1/2 = 1 {V6 — V2). 

This is also the value of cos 75°, since 75° + 15° = 90°. 
2. Find the values of cos (45° d= ^). 
In (39) and (41) put X = 45°, 2/ = ^. 



cos (45"^ 
cos (45^ 



0) = cos 45° cos d — sin 45° sin 6, 
e) = cos 45° cos + sin 45° sin 6. 



But 



cos 45° = sin 45' 



t/2. 



. •. cos (45° ±zd) = iV2 (cos d =F sin 



3. Show that cos 15° 

4. Show that sin (45° ± 

5. Show that 

(a) sin (30° 

(b) cos (30° 



i{V6 + 1/2) == sin 75°. 
:d) = ^ v"'2 (cos d d= sin 8). 



0) 



0) 



cos 6 rh 1/3 sin 



1/3 



cos d ^ sin 



6. Prove some of the formulse (18), (20), (22), (24), (26), (28) by 
using (38), (39), (40) and (41). By this means these formulae are 
shown to be true for all values of 6, as was stated, § 22, note. 

30. The Tangent and Cotangent of the Sum and Dif- 
ference of Two Angles. Dividing (38) by (39) we have 

sin X cos y -\- cos x sin y 

tan (x -\-y)= ^-^—. ^. 

^ "^^ cos 0.' cos 3/ — sm a; sm 2/ 

Dividing both terms of the fraction by cos x cos y, 



GENEEAL FOEMUL^. 53 



Sin X cos y cos x sm y 

cos X cos 'v cos X cos 7/ 

tan (x + y)= ^ , ^^. 

^ ^^ cos a:^ cos 2/ sm .T sm 2/ 

cos oj cos 3/ cos X cos 2/ 

tana? -1- tan 2/ ,,^- 

tan(. + ^) = --^^±^. (42) 

In like manner, from (40) and (41) 

sin X cos y — cos x sin y 

tan (oj - ^) = ^-^ ^, 

^ "^^ cos a; cos 2/ + sm ic sm y 

from which, as above, 

tana? — tan?/ 

^"^^"-^-^)= l + tan^tan.y ^4^) 

The student can also prove in an analogous manner that 

cot a? cot ?/ — 1 

cot (x + y) = — — , (44) 

^ ^ cot ^ + cot X ^ ^ 

cota?coti/+ 1 

and cot (a? — :?/) = — — . (45) 

^ ^ cot 2/ — cot a? ^ ^ 

Exercises. 1. Find the value of tan {6 + 45°). 
In (42) put a; = ^, 2/ = 45°. Then 

^ ,^ , .^^, tan ^+ tan 45° 

^""^^^'') = l-tan.tan45° ' 

But tan 45° = 1, hence 

/^ , ^^nx 1 + tan 6 

tan (6 + 45°) = -^- . 

^ ^ ^ 1 — tan^ 

2. Show that tan {B - 45°) = ^^" ^ 7 ! • 

^ ^ tan ^ + 1 

o r^n \li ^ sin (x + y) tan a; + tan y cot v + cot x 

3. Show that . ) i = r ^^^^ = J r"- 

sm (a; — y) tan x — tan y cot y — cot x 

. ^, , , cos (x 4- y) 1 — tan x tan i/ cot x cot w — 1 

4. Show that ) — ^ = — — ^ = ~- / , ., ♦ 

cos (x — y) 1 -\- tan x tan ?/ cot x cot 2/ + 1 

5. Find the values of tan and cot of 15° and 75°. 



54 A SHORT COURSE IN TRIGONOMETRY. 

31. Functions of Double Angles. The following for- 
mulae express the values of sin, cos and tan of twice an angle 
in terms of functions of the angle. They are derived by 
making y = xm (88), (39) and (42). Thus from (38), we 

have sin {x -]- x) = sin x cos x -f cos x sin x, 

or- sin 2a5 = 2 sin a? cos x, (46) 

In like manner, from (39) and (42) 

cos 23C = cos^ Qc — sin^ a?, (47) 

2 tan oc 
*^''2-=i-t,^. (48) 

Exercises. 1. Derive the values of sin, cos, and tan of 60°, 
from the functions of 30°. 

From (46) sin 60° = 2 sin 30° cos 30° = 2 • | • J i/3 = J i/3. 
From (47) cos 60° = cos^ 30° — sin^ 30° = | — ^ = J. 

2 tan 30° 1 1/ 3 f i/ 3 ,- 
From (48) tan 60° = ^^^^^^^. = 1^ = i^ = / 3. 

2. From the values of sin, cos and tan of 75° given in the 
results to exercises 1 and 3, § 29, and exercise 5, § 30, find the 
values of sin, cos and tan of 150°. 

3. Given tan ^ = 3. Find sin, cos and tan of 26, and construct 
both angles. 

4. Show that sin 3a: = 3 sin x — 4 sin^ x. 

Note : Put y = 2xin (38), and in the result substitute for sin 2x 
and cos 2a;, their values from (46) and (47). In this result sub- 
stitute 1 — sin^ x for cos^ x and collect terms. 

5. Show that cos 3a; = 4 cos^ a; — 3 cos x. 

3 tan X — tan^ x 



6. Show that tan 3a: 



3 tan^ X 



32. Functions of the Half-Angle. The following for- 
mulae express the values of sin, cos and tan of half an angle 
in terms of the cos of the angle. 



GENERAL FORMULA. 55 



From (47) 


cos^ ^x — sin^ ^x = cos x. 


and from (4) 


cos^ ^x + sin^ ^x = 1. 


Subtracting 


2 sin^ -oc= 1 — cos oo. 
2 


Adding 


2 cos^ -x= 1 + cos a?. 


Dividing (49) by (50) 



(49) 
(50) 



1 1-cosa? 

tan^ -x = ~— . (51) 

2 1 + cos a? ^ '' 



These results may also be expressed thus 



sin 2 



1 /I — cos X ._„^ 



cos 



1 ll + coscc ,^^^ 

2'^=V^^' <5^) 

1 ll — cosa? 

Exercises. 1. From the functions of 180° derive those of 90°. 
From (52) sin 90" = ^"[E^J!^ = V| = l" 
From (53) cos 90» = J^+^^l = J| = 0. 



■r. /K^N + AAO 1 — COS 180° 2 

From (54) tan 90° = \- — - — -\~ = 

^ ^l^- cos 180° ^0 



Note : The operation above appears to give t/ 1 = zh 1 as the 
value of sin 90°. Only the positive value is written, as it is 
known that sin 90° is positive. The explanation lies in the fact 
that cos 540° has the same value as cos 180°, and hence the re- 
sult gives sin 270° as well as sin 90°. 

2. In (49) write 90° — ^ for x. Then 

2 sin^ 1(90° — ^) = 1 — cos (90° — 6)^ 
4 



56 A SHORT COURSE IN TRIGONOMETRY. 

or 2 sin^ (45° — ^d) = 1 — sin 6. 

Now(45°— J^) + (45°+i^)=90°,.-. sin(45°— i^)=.cos(45°+J^). 
Hence we have also 

2 cos^ (45° + 1^) = 1 — sin 6. 

3. Given cos 45° = ^V 2, find the values of sin, cos and tan of 
22° 30^ 

4. Given cos 6 = ^ find sin, cos and tan of ^6. 

5. Given sin B = — |, find sin, cos and tan of Id. 

6. Show that 

2 sin^ (45° -\-^e) = 2 cos^ (45 — J^) = 1 + sin 6. • 

33. Sum and Difference of the Sines and Cosines of 
two Angles. Take the sum and difference of (38) and (40), 
and of (39) and (41). Then 

sin (x -{- y) -\- sin (x — y) = 2 sin x cos y, 
sin (x -\- y) — sin [x — y') = 2 cos x sin y, 
cos (x -^ y) -{- cos (x — y) = 2 cos x cos y, 
cos (x -\- y) — cos (x — y) = — 2 sinx sin y. 

Let X -\- y = o.y and x — y = j^, 

then ic = 1 (a + /?), and y = ^ (a — /?), 

Substituting these vahies above, we have 

sin a + sin /8 = 2 sin 1 (a + /3) cos J (a — yS), (55) 
sin a — sin y8 = 2 cos 1 (a + yS) sin J (<x — /3), (56) 
cos a H- cos y8 = 2 cos 1 (a + /3) cos J (a — /S), (57) 
cos a — cos /8 = — 2 sin J (a + y5) sin J (a — 13), (58) 

Dividing (55) by (56) 

sin a + sin /9 ^ , ^^ , , ., 

-^ . ^ = tan J (a + /5) cot i (a - B\ 

sm a — sm /? ^ ^ ^^ ^ ^ ^^^ 

sin a + sin ^ tan i (a + ^8) 

sm a — sin p tan f [a — p) ^ ^ 



GENERAL FORMULA. 57 

Exercises. 1. Show that 

(sin 74° + sin 16°)(cos 53° — cos 37°) = — 2 sin 8° cos 29°. 

By (55) sin 74° + sin 16° = 2 sin 45° cos 29°. 

By (58) cos 53° — cos 37° = — 2 sin 45° sin 8°. 

. •. (sin 74°4-sin 16°)(cos 53°— cos 37°) =— 4 sin^ 45° sin 8° cos 29°, 

= — 2 sin 8° cos 29°. 

2. Show that cos ^ = 2 cos (45° + ^d) cos (45° — J 6). 
The expression on the right is the same as 

2 cos 1 (90° + 6) cos i (90° — 6), 

which by (57) equals cos 90° 4- cos 6, but cos 90° = 0, hence this 
last expression equals cos 6. Therefore the given equation is 
proved. The same thing may be proved thus : 

2 cos (45° -i-ie) cos (45° — J ^) = 2 sin (45° — ^d) cos (45° — J 6), 

= sin (90° — d), by (46), 

= cos 6. 

3. Show that sin 5a — sin 3a = 2 cos 4a sin a. 

4. Show that ^Q^ a - cos ^ _ _ tan 1 (« + (^) tan J (a-/5). 

cos a + cos /3 ^ ^ ^ ^ 

5. Show that cos (30° — 6) ~ cos (30° -^6) = sin d. 

Examples. 
In the following examples prove the identities given, except 
when otherwise stated. 

- tan a tan /? _ sin a sin /3 

tan a — tan j3 sin (a — /5)' 

2. tan x = 2, tan y = j, find sin, cos and tan of (x ± y). 

(a) If X and y are both in the first quadrant. 

(b) If X is in the third quadrant and y in the first. 

« cot X + tan V / s 

o. — 7 = cos (x — y) sec (x + w). 

cot x — tany ^ ^' ^ ' ^^ 

4. sin (a 4- /5) cos a — cos (a + /5) sin a = sin /?. 

5. sin (a — /5) cos /? + cos (a — /?) sin ^ = sin a. 

6. cos nQ cos 4- sin nQ sin ^ = cos {n — 1)^. 

7. sin nd cos (n — 1)^ — cos nd sin (?i — 1)^ = sin ^. 



58 



A SHOET COURSE IN TRIGONOMETRY. 



8. 

9. 
10. 
11. 

12. 

13. 



T^ I 1 nno I. XT- X X 1 — tan a^ tan v 

Itx-^y-\-z = 90°, show that tan z = ". 

tan X + tan y 
c os{x~\- y ) _ 1 — tan x tan y 
sin {x — y) tan x — tan y ' 

cos X sin {y — z) -\- cos y sin (z — x) -{- cos z sin {x — y) = 0. 
cos^ a 4- sin^ a __ 
cos a 4- sin a 



sm^a 



14. tan 



cos a — Sm a 

cos(^+45°)_ 
cos [d — 45°) ~ 

sin ^ 



J sm 2a. 
J sin 2a. 



= 1 

= sec 20 — tan 26. 



sin 20 



15. 
16. 



1 + cos + COS 26 
tana;=l, tan 2/ = i^, find tan (a; + 2i/). 
2 sin 2a: = (sin x + cos xy — (sin a: — cos xy. 



CSC^ 



csc^0 



17. {a) cos 20 

18. esc 26 = h sec esc 0. 



(6) sin 20 



2 tan 



19. 



cos a 



1 + tan 
sin a 



tan — X- 

2i 



COS a; 



sm X 1 + cos X 

1 4- cot X cot \x = CSC X cot ^a;. 



20 

22 

23. tan H« + /3) — tan H« — /5) 

24. 



cos a — sm a 
sin X 



21. cosa; = 



2 sin/? 



= sec 2a + tan 2a. 
1 — tan^ ^x 



1 + tan'-^ ^x 



cos a + COS /5 



/ l + tani0 Y_14- 
\1 — tani0/ ~1 — 



sin 



sin 0' 



25. 
26. 
27. 

28. 

29. 



{a) sin 
sin fa; = 
(a) sin a 



cot J0 + tan J0 

(1 + 2 cos 0) sin |0. 

1 



~; (5)tan0 = 



cot -|0 — tan J0 



tan^ ~ X = 



cos 



tan J a + cot a 
2 sin a: — sin 2a; 
2 sin a; -f sin 2a;* 

2 



; {b) cos a = 



cot a 



tan Ja + cot a 



n 



tan (45° + |0) + tan (45° — |0) 



GENERAL FORMULiE. 59 

30. tan e = tan (45° + ^>^)- tan (45°-^^) 

_, sin a; + sin w , , , , . 

31. \- ^ = tan i(a; + I/). 

cosx + COS 2/ 

-- sin X — sin 1/ ,. . . 

32. ^ = tan l(x — y), 

cos X + cos 2/ 

-_ sin X ^ siny 

33. — - = — cot Ux — v). 

cosx — cos^ ^ "^^ 

- . sin X — sin y ^ , , , . 

34. = — cot Ux + y). 

cos a; — cos y ^ \ ^7/ 

35. cos (60° + 6/) + cos (60° — <9) = cos<9. 

36. cos d + cos 3^ + cos 5^ + cos 76 = 4: cos 6 cos 26 cos 4^. 

cos a — cos 5a 



37. Simplify 

38. Simplify 



sin a -\- sin 5a 
sin 5a — sin 3a 

cos 5a + COS 3a* 



„. sin Cn — 2)a + sin na 

39. ) ^^ = cot a. 

COS {n — 2) a — cos na 

cos na — cos (n + 2)a 

40. -^ — ^ — -^rr , — ~ = tan (n + l)a. 

sm ('/I + 2)tt — sm na \ > / 

41. tan 50° + cot 50° = 2 sec 10°. 

.- cos (a; — Bv) — cos (3a; — y) ^ . , 

42. ^ — . ^^ , — ^ ^ = 2sm(a; — v). 

sm 2a; + sm 2y ^ ^^ 

43. 2 sin (a + 45°) sin (a — 45°) = sin^ a — cos^ a. 

1 + 2 sin a _ tan (15° + Ja) 
' 1 — 2 sin a "~ tan (15° — Ja)* 

45. 2 CSC 4^ + 2 cot 46 = coi6 — tan 6. 

46. cos 4^ = 8 cos* ^ — 8 cos^ ^ + 1. 

. _ 3 sin X — sin 3a: ^ „ 

*7. — 5 — r^ = tan^ x. 

cos 3a; + 3 cos x 

._ ^ ,. sin 2^ cos ^ 

48. tan 16 = 



(1 + cos 2^)(1 4- cos 6) 



CHAPTER Y. 



SOLUTION OF OBLIQUE TRIANGLES. 

34. Theorems. 

I. In any triangle the sides are proportional to the sines of 
the opposite angles. 

C 





A C D B AC 

Fig. 30. 



B D 



Let ABC he any triangle. Let fall a perpendicular from 
(7 to AB. Then in both figures 

p 



and 



sin A = J-, or p = b sin A. 



sin B = —, or p = asm B: 

a' -^ 



a sin B = h sin A. 
Dividing by sin A sin B we have 

a h 



% 





sin 


A 


sin B' 




r way 








1 




h 


c 




and 


G 


a 


sin B 


sin C 


sin C~ 


sin A' 




a 




b 


c 





sin A sin B sin C * 

60 



(60) 



SOLUTION OF OBLIQUE TRIANGLES. 61 

II. In any triangle the sum of two sides is to their differ- 
ence as the tangent of half the sum of the opposite angles is to 
the tangent of half their difference. 

From ,- = -. — ^ bv composition and division, 
6 sm ^ " ^ 

a -\- b sin J. + sin 5 
a — 6 ~ sin J. — sin B' 

But, by (59), sin A + sin B tan j (A + B) ^ 
sin A — sin B "~ tan ^(A — B)' 



a — &"" tani(^ — jB)' 



(61) 



with similar expressions for the other pairs of sides. 

III. In any triangle the square of any side is equal to the 
sum of the squares of the other tivo sides minus twice the con- 
tinued product of these two sides and the cosine of the included 
angle. 

First, when the side is opposite to an acute angle. 

In the first triangle, Fig. 30, BD = AB — AD, and in 
the second, BD = AD —AB ; therefore in both cases 

BD' = AB'-j-AD'-2ABAD. 

To both sides of this equation add CD^, then 

' BD' + CD' = AB' + AD' +CD'- 2ABAD; 

but BD' +CD' = BC', AD' -\-CD'=: AC', 

and AD = AC- cos A: 

BG' = AB' + AC' - 2AB • A (7- cos A, 

or a' =b' -{- c' — 2bc cos A. 

Secondly, when the side is opposite to an obtuse angle. 



62 A SHORT COURSE IN TRIGONOMETRY. 

C AI) = AB-\-BD; 




Adding CD- to both sides, we have 
AI)'-i-CI>' = 

AB' + Bl)'+ CD' -}- 2AB BJD. 

But AD' +CD' = h\ BD' + CD' = a', 

and BD = BC cos (180° - 5) =_ a cos B, 

Hence b' = c' -{- a' — 2ca cos B. 



It follows therefore that in any triangle 

a^=b' -}- c' — 2b€C0sA, 
b^ = c' -\- a' — 2ca cos B, 
c2 = ^2 _j_ ^2 _ 2ab cos C. 

35. The Sines of the Half-angles of a Triangle. 

From the first of (62) we have, solving for cos A, 



(62) 



cos A — 



26c 



Subtracting both sides of this equation from unity, 



cos A= 1 



26c -b'-e'-j-a' 



26c 



(6-c)^ 



26c 26c 

(a + 6 — c) (a — 6 4- c) 
"" 26c 

. .r.. . o , A (a -}- 6 — c) (a — 6 + c) 
by (49), sm'iA = ^-^ ^^ ^-1 



Now, let 
Subtracting 



a -{- b -h c= 2s 

2c = 2c 
a + 6 — c = 2 (.s — c). 




SOLUTION OF OBLIQUE TRIANGLES. 



63 



Likewise, 



Also, 



a — h -\- G= 2(s — h). 
(s-b){s-c) 



sin^ J A 



be 



• 9 -. ^ (s — c)(s — a) 



ca 



sin' i C = 



(« — a) (s — &) 



a& 



(63) 



36. The Cosines of the Half-angles of a Triangle. 

cos' 1^ = 1 _ sin' 1^ ; by (4) 

5c — (s' — bs — cs -\- bo) 
^ bF" 

s(b -\- G — s) 



cos^ 



by (63) 



But 

Hence 

Also, 



be 



6 -j- c = 2s — a, .'.b -j- G — s = s — a. 



cos' ^ A = 



cos' 1 ^ 



cos' J C =^ 



s (s — a) 
&c 

s (« — 6) 

s (s — c) 
ah 



(64) 



These formulae may also be derived like those of § 35 by 
starting with the equation, 



cos J. = 



6' + c' - a' 
26c 



and adding both sides to unity, reducing and using (50). 
The student should derive them in this way. 



64 



A SHORT COURSE IN TRIGONOMETRY. 



37. The Tangents of the Half-angles of a Triangle. 

If we divide each one of (63) by the corresponding one of 
(64), we obtain 



tan^ 1 A 



(s -b)(s~ e) 
s (^s — a) 



„ , (s — c) (s — a) 
tan2 1 J5 = ^ y^ — -^ 

^ s(s — b) 



tan^ 1 C = 



(s — a) (^s — h) 
s (s — c) 



(65) 



If we write 



= aI^— 



){^-b){B-c) 



then since tan^ J ^ = 



{s — a) (s — h) (s — c) 
s(s — of ' 



we have 



or 



similarly 



tan2i^_^^_^^,, 



tan J^ 



— a 



tan \B— , 



tanjC 



s — h 

r 
s — c 



n 



(66) 



To determine the angles of a triangle when the sides are 
given (66) are more convenient than (65). 

The next step is to show how the formulae derived in this 
chapter are applied to the solution of oblique triangles. 
This problem will be divided into four cases, as follows : 



* r is the radius of the inscribed circle of the triangle, as will be shown 
in § 46. 



SOLUTION OF OBLIQUE TRIANGLES. 65 

Case I. When a side and two angles are given. 

Case II. When two sides and the angle opposite to one 
of them are given. 

Case III. When two sides and the included angle are 
given. 

Case IV. When the three sides are given. 

38. Case I. A Side and Two Angles Given. Let the 
given parts be Aj By and a. 

The formulae are (7= 180° - (^ + B), and from (60) 

a sin 5 a sin C 

sm A sin A 

For a check compute the required angle from the three 
sides by the appropriate one of (63), (64), or (65). 

Example: ^ = 47° 19^, 5 ^78° 15^, a = 738.2. Find C, h 
and c. 

Solution. Check. Use the last of (65). 

A^B= 125° 34^, C= 54° 26^ s = K« + ^ + c) = 1269.0 

s — a= 530.8 
log a = 2.8682 s — h= 286.0 

log sin J5 = 9.9908 s — c= 452.2 

cologsin (7 = 0.1336 

log h = 2.9926, h = 983.0 colog s = 6.8965 

log (s — a) = 2.7249 

log a = 2. 8682 log {s — h) = 2. 4564 

log sill C = 9. 9103 colog (s — c) = 7. 3447 

colog sin A = 0.1336 log tan^ iC= 9.4225 

log c = 2.9121, c = 816.8 log tan ^C= 9.7112 

C=54°26^, JC=27°13^ 

Examples. 
Solve the following triangles : 

1. A= 26° 19^ B= 58° 29^ 6 = 6.382. 

2. 5 = 102°38^ C= 42° 37^ a = 26.47. 
Z. A= 64° 37^ (7= 37° 2r, 6 = 562.9. 
4. ^= 51° 42^ C=119° 8^ c= 7.617. 



66 



A SHORT COURSE IN TRIGONOMETRY. 




39. Case II. Two Sides Given, and the Angle Oppo- 
site to One of Them. Let the given parts be a, b and A, 

If the given angle is acute, and the side opposite this angle 
is less than the other given side there will be in general two 
solutions. A figure will make this clear. Draw the given 
angle A, Fig. 32, and lay oif a dis- 
tance b on one of its sides. This 
will establish the vertex C. With b 

Cas a center, and a radius equal 
to a describe the arc B^B^. Then 
both triangles AB^Q AB^C fulfil 
the given conditions. 

Should a be just long enough to reach from C to AB, that 
is if a = b sin A, there will be only one solution, a right tri- 
angle, ABC. If a < 6 sin A, 
it is evident that no solution 
is possible. 

If A is acute, and a^ b, 

there will be but one solution. 

For repeating the construction 

described above the result is 

^^^' ^^- as shown in Fig. 33 ; where 

the triangle AB^C does not fulfil q 

the given conditions, so that AB^C 

is the only solution. 

If the given angle A is obtuse 
there is one solution if a > 6, and 
none if a<6, as appears from Fig. 34. 

Note : While it is always possible to determine beforehand, in 
the manner just described, whether the solution will be ambiguous 
or not, the same thing will come out in the course of the arith- 
metical solution. 

The formulse for this case are 
. ^ b sin A ^ . ^x « sin C 





SOLUTION OF OBLIQUE TRIANGLES. 67 

the first and last being derived from (60). For a check one 
of the angle may be computed from the three sides, as in the 
check to Case I. 

Example! ^ = 38 ° 55^, a = 625. 9, 6 = 739. 5. Find B, C and c. 

log 5 = 2.8689 B^ = 4:7°55', B^ = 1S2° 5' 

log sin A = 9.7981 ^ + j5, = 86° 50^, A + B.^ = 17r 0' 

cologa = 7^035 C=93°10^ C,= 9° 0^ 

log sin B = 9:8705* 

log a = 2. 7965 log a = 2. 7965 

log sin C, = 9.9993 log sin C, = 9.1943 

colog sin ^ -= 0. 2019 colog sin A = 0.2019 

log c, = 2.9977, Cj = 994.8j log c^ = 2.1927, c^^ 155.9, 

The three given parts form one triangle with B^, C^ and Cj, and 
another triangle with B^, C^ and c^. 

For a check formula use the last of (65). The logarithmic work 
may be arranged as on page 65. 

Examples. 
Solve the following triangles : 

1. A= 60° 36^, 5 = 62.19, a = 75.73. 

2. 0= 37° 41^, 6 = 1.963, c = 1.728. 

3. 5 = 105° 38^, a = 238.7, 6 = 624.3. 
4.5= 59° 27^ c= 26.12, 6 = 22.62. 

5. C= 12° 42^, a = 123.9, c = 201.8. 

6. A= 43° y, c= 52.98, a = 47.63. 

40. Case III. Given two sides and the included angle. 

Let the given parts be a, b and C. The formulae for this 
case are A -{- B ^ 180° - C. 

tan i{A -B)= ^^^ tan ^{A + B), from (61 ) 

A = i{A + B) + i(A-Bl B:=i{A-hB)-i{A-B) 
a sin (7 b sin C 






sm A ' sm B 



* Whenever an angle is found from its sine, there will always be two 
results less than 180°, since the sine of an angle is equal to the sine of its 
supplement. See (18). 



68 A SHORT COURSE IN TRIGONOMETRY. 

The agreement of the values of c as computed by the last 
two formulae will serve to show the accuracy of the work. 

In the formulae written above a is supposed to be larger 
than h. If the reverse is true write h — a, and ^ — A. 

Example : A = 49.39, h = 41.41, C= 109° 38^ 

a — h = 7.98, a + 5 = 90.80, A-\-B = 180° — C= 70° 22^ 

log {a — h) = 0.9020 ^{A -\-B) = 35° 11^ 

colog(a+5) = 8.0419 1{A — B)^ 3° 32^7 
log tan ^{A + B) = 9.8482 A = 38° 43^7 

log tan K^ — 5) = 8. 7921 5 = 31° 38^.3 

log a = 1.6936 * log5 = 1.6171 

log sin C= 9.9740 log sin C= 9.9740 

colog sin A = 0.2037 colog sin B = 0.2802 

log c = 1.8713, log c= 1.8713, c = 74.35. 

41. Special Solution for Case III. Let the given parts 
be a, b, and C. Take an auxiliary angle x such that 

tan ^ = T, where a^h, 

T T . . tan X —\ a — b 

then by composition and division, — r = — — -^ ; but by 

tan X -J- 1 ct -f- 

tan f — 1 
^^- ^^ § ^^ t"a^¥Tl = *^^ (^ - ^^')^ 

^^^ = tan(aj — 45°). 

If this value be substituted in the formula used for com- 
puting tan ^ {A — B)yWe have 

tan ^ {A- B) = tan (x - 45°) tan J (A -f B). 

The solution is then completed as before. 

Note : There is a slight saving of labor in this method, be- 
cause we avoid the necessity of finding a + 6 and a — b and of 




SOLUTION OF OBLIQUE TRIANGLES. 69 

looking up their logarithms. Logarithms of a and h are found at 
the start, and are then ready when required at the end for the 
computation of c. If the triangle forms one of a series, and log- 
arithms of a and h have been found by previous computation, 
they can be used directly in these formulae without finding the 
actual values of a and h. It is desirable, in the formula for tan x, 
to place the larger of the given sides in the numerator, so that 
tan X shall be greater than 1, and hence x greater than 45°. 



Example: a = 502.9, 5 = 326.8, C=78°16^ 


log a = 2.7015 


0; = 56° 59^3 


log & = 2.5142 


a; — 45° = 11° 59^3 


log tan re = 0.1873 


1{A + 5) = 50° 52^ 


log tan {x — 45°) = 9.3271 


iIa — B) = 14:°Z1\Q 


log tan \{A -f B) = 0.0895 


^ = 65° 29^6 


log tan ^{A — B) = 9.4166 


5 = 36° 14^4 


log a = 2.7015 


log 6 = 2.5142 


log sin C= 9.9908 


log sin C= 9.9908 


colog sin ^ = 0.0410 


colog sin 5 = 0.2283 


log = 2.7333, 


log c = 2.7333, c = 541.1 


Examples. 


Solve the following triangles : 




1. a =72.83, 6=38.16, C 


= 61° 16^ 


2. Z> = 1.612, c = 1.796, A 


= 112°48^ 


3. c =632.9, a = 217.3, B 


= 29° 32^ 


4. a =4.718, 6 = 6.312, C 


= 96° 12^ 


5. 6 = 59.63, c= 38.96, A 


= 74° 24^ 


6. c= 269.8, a = 347.5, B 


= 56° 38^ 



42. Given Two Sides and the Included Angle to 
Find the Third Side directly from the Data. 

Let a, h and (7 be given. Then by (62) 

c2 = a2^62_2a5cosC. 

To change this formula to a form suitable for logarithmic 
computation add and subtract lah on the right, then 

G^=.o?- 2ah + ¥ + 2ah - 2ah cos C 



1 



70 A SHORT COUESE IN TRIGONOMETRY. 

= (a - hf + 2a6 (1 - cos C) 

= {a- by + 4a6 sin^ JC. by (49) 

c^ ^ 4ab sin^ J C 

Hence 7 ^, ^3 = 1 + ~7 r\2~ • 

(a — 6)^ (a — 6)^ 

Now take an auxiliary angle d, such that 

^ „ _ 4a6 sin^ 1(7 

tan^ /? = — j^^ , 

(a — 6)^ ' 

(.2 

then 7 pro =14- tan^ 6 = sec^ ^ = ^ ^, 

(a — by cos^ <?^ 

from which c = ^. fw) 

cos ^ ^ ^ 

By means of (i) and (ii) the value of c may be found. 

Examples : 1. Given a = 123.4, b = 87.6,C= 54°. Find c by 
(i) and (ii) above. 

log 4 = 0.6021 a — 6 = 35.8 

loga = 2.0913 ^=69° 14^ 
log 6 = 1.9425 

log sin^ 1 C = 9.3140 log {a — b) = 1.5539 

colog {a — by = 6.8922 log cos ^ = 9. 5497 

log tan^ 6 = 0.8421 log c = 2.0042 

log tan ^ = 0.4211 c = 101.0. 

For a check operation compute JC by the third of (65). 
2. In each of the examples at the end of § 41 compute the un- 
known side directly from the data. 

43. Given the Three Sides. The three angles can be 
found from either (63), (64), (65), or (66). The last give 
the solution in most concise form, and are in other respects 
the best. The results obtained from (63) are unsatisfactory 
if one of the half-angles is nearly 90°, and (64) are equally 




SOLUTION OF OBLIQUE TKIANGLES. 71 

undesirable if one or more of the half-angles is very small.* 
The tangent fornmlse are not open to these objections. 
Hence we use 



= 1 (a + 6 + c), r = ^ 



(s — a) (.s ~b)(s — c) 



/y ^ /yt 

tan ^A = , tan ^B — r , tan i (7 = 



2 ^-L — ^ ^ , •-"" 2-^ — o __ i)> '^"'" 2 



S — G 



If the sum of the three half-angles, as determined, equals 90°, 
the work is right. 

Example: a = 16.92, b = 21.43, c = 19.07. 

a = 16.92 cologs = 8.5420 ^A = 24:°2d' 

6=21.43 log (s — a) = 1.0715 ^5 = 36° 24^ 

c = 19.07 log (s — b) = 0.8621 JO = 29° 7 ' 

2s = WA2 log (s — c) = 09841 90° 0^ 

s = 28.71 log 7-2 = 1.4597 

s— a =11.79 log r = 0.7298 ^ = 48° 58^ 

s—b= 7.28 log tan J^ = 9.6583 .B = 72° 48^ 

s-c= 9.64 log tan J5= 9.8677 = 58° 14^ 

log tan 1(7 = 9.7457 

Examples. 
Solve the following triangles. 

1. a = 329, b = 627, c = 578. 

2. a = 1.623, b = 1.798, c = 1.021. 

3. a = 14.92, 6 = 23.16, f? = 19.36. 
■ 4. a = 4263, b = 5984, c = 6217. 

44. Exercises and Applications. Many different types of 
problems depend for their solution upon the trigonometric solu- 
tion of triangles. A few of these types are illustrated in the 
problems worked below. In attacking such a problem it should 
first be carefully analyzed, and the method of solution outlined, 

* This arises from the fact that with angles near 90° the sine changes 
very slowly, and hence such angles cannot be determined with great pre- 
cision from the sine. Similarly with small angles the cosine changes 
slowly, and hence such angles cannot be determined with precision from 
the cosine. 



72 



A SHORT COURSE IN TRIGONOMETRY. 



before any of the arithmetical work is performed. Sometimes a 
general solution can be obtained ; that is, denoting the given 
quantities by letters, the quantity or quantities required can be 
expressed directly in terms of the given quantities. This is the 
most instructive method and should be adopted whenever possible. 

1. Two points A and B are on 
opposite sides of a stream, but on 
account of an island in the stream 
B is invisible from A. A straight 
line, CD, passing through A is meas- 
ured along the bank. CA = 496 feet, 
AD =72S feet, BCA = 41° 12' and 
ADB = 47° 38^, determine AB. 

Since Ci>, and the angles C and D are known, the triangle 
CBD can be solved. Then AB can be found from CB, CA and C, 
or from BD, AD and D. In fact AB should be computed both 
ways to ensure the accuracy of the result. The formulae are 




BC = 



CBD = B= 180^ 
CD sin D 



sin^ 



BD 



(C + D). 
CD sin C 
sin B^' 



by (60) 



tan2 e. 



tan2 e, = 



4:ACBCsm^G 

{BC—ACy ' 

JADBDsm^D 
{BD — ADf ' 



AB = 



AB = 



BC-AC 

cos d^ ' 

BD- AD 



COS^o 



by(*) 

and (n) 
H2. 



91°10^ 



log CD = 3.0860 
log sin C= 9.8187 
cologsin 5 = 0.0001 
log BD = 2.9048 
J5i> = 803.2 
^i> = 723.0 
|C=20°36^ 50— J[ (7=405.0 ID=2^°4^\ BD—AD = 80.2 



log CD = 3.0860 

log sin i) = 9.8686 

colog sin 5 = 0.0001 

log 5C= 2.9547 

5C= 901.0 

^(7=496.0 



log 4 = 0.6021 

log CA = 2.6955 

log C5 = 2.9547 

logsin^ I- C= 9.0926 

colog (5C — ACf = 4.7850 

log tan^ d, = 0.1299 

log tan 6^^ = 0.0650 



log 4 = 0.6021 

log ^i>= 2.8591 

log 5i) = 2.9048 

logsin^ ii)= 9.2124 

colog {BD — ADf = 6.1916 

log tan^ ^2 = 1-7700 

log tan ^, = 0.8850 



SOLUTION OF OBLIQUE TRIANGLES. 



73 



e^ = 49° 16^ 

(5C-^C) = 2.6075 

log cos 0^ = 9.8146 


^2-=82°34i 

log {BD — AD) = 1.9042 

log cos ^2 = 9-1113 


log AB = 2.7929 
AB = 620.7 


log ^5 = 2.7929 
^5 = 620.7 



2. To find the altitude of a point P above a horizontal plane 
two points A and B are taken in the horizontal plane and 
in the same vertical plane with 
P. The angles of elevation of 
P at A and B are 36° 38^ and 
25° 16^, respectively, and AB = 
600 feet. What is the altitude re- 
quired ? 

Let the given angles PAK and 
PBK be denoted by a and j3 re 
spectively, and let AB = a. Then APB 




Fig. 36. 
- (3. Hence 



AP = 



AB sin PBA 
sin APB 



and 



PK = AP sin a = 



sin (a — /?)' 
a sin a sin /? 



Using the numerical 
values given above for 
a, a, and (3 the result 
is obtained as shown 
on the right. 



PK 



sin (a — 13) 

loga= 2.7782 

log sin a= 9.7758 

log sin /3 = 9.6302 

colog sin {a — (3) = 0.7053 

775.3 feet. log PK= 2:8895 



3. A line AB is tangent to a circle at A, and a secant through 
A cuts the circle in C and D. The angle between the secant and 
tangent is 15° 38^, BC=67, and BD = 73. Find the length of 
the radius of the circle and the lengths of the chords AC and BD. 

From geometry we know that AB 
is the mean proportional between BC 
and BD. Hence triangle ^^C can be 
solved. For convenience denote the 
angles CAB, ABC, ACB by a, (3, y re- 
spectively. Then AC and AD are 
computed from the triangle ADC, in 
which CD= BD — BC, ACD = 180° 
— y, and ADC = B AC = a. Draw the 
radii OA, OC, OD, and draw OE, Oi^ bisecting ^OC and AOD re- 




74 



A SHOET COUESE IN TEIGONOMETEY. 



spectively. Then AOE=ADC=a^ AOF = ACD = 180° — y, 
A g and the radius, r, is computed from 

either AOF or AOE. 

Formulse : AB = VBCBD ; 
tan Ky - «) = .p , pr r^an |(y + a), 




^^ + BC 



AC = 
BC sin ^ 



AB sin 



AD = 



sm 7 

i>C sin 7 



^ C sin 7 



^X> 



2 sm a ' sm a sm (7 — a) 2 sm 7 

The student should supply the logarithmic work. The results 
are ^0=18.12, AD = 20.51, and r = 10.69. 

4. A, Bj and Care three points 
on a straight line. AB — a, and j^, 
BC^= b. P is a point such that 
the angles APB and BPC are 
equal. Find the distance AP. 

Denote the equal angles A PC 
and BPC by a. Since PB bisects 
ABCj we have 

PC _h_ 
PA~ a' 



Also 
Hence 



tan 1{A —C) b—a' *" ' ^ 

or, since ^{A -j- C) = 90° — a, tan ^{A + C) = cot a, 

Z> — a 




Fig. 38. 



PC sin A sin ^ 


b 


P^~sinC' •• sinC" 


a 


sin ^ + sin C b ^ a 




sin ^ — sin C b — a 




tan ^{A+C)_b-ha 


59 



and 



tan h{A — C) 



b + a 
Then ^ and C are known and ABP = C -^ a 



cot a. 



hence 



AP = 



a sin ABP 



and 



^P 



(a + b) sin C 



sm a ' sin 2a 

Suppose a = 624 feet, 6 = 900 feet, and a = 37°. 



^ 



SOLUTION OF OBLIQUE TEIANGLES. 75 

log {b-a) = 2.4409 . ^{A + C) = 53° 

colog (6 + a) = 6.8170 ^{A — C) -= 13°3r 



log cot a ==0.1229 


A = 66°SV 


an J(^- (7) = 9.3808 


C=S9°29' 




ABP=a + C=76°29' 


log a = 2.7952 




log sin ^^P= 9.9873 


log (aH- 6) = 3.1830 


colog sin a = 0.2205 


log sin C= 9.8033 


log AP = 3 0035 


colog sin 2a = 0.0172 



AP = 1008 feet log AP = 3. 0035 

Examples. 

1. The longer diagonal of a parallelogram is 500 feet, and the 
angles it makes with the sides are 46° 36^, and 10° 12^. Find the 
lengths of the sides and of the other diagonal, and the angles the 
latter makes with the sides. 

2. The length of a diagonal of an isosceles trapezoid is 237 feet. 
At one end the angles it makes with the base and side are 18° 12'' 
and 37° 27^, respectively. Find the lengths of the four sides. 

3. The lengths of the bases of a trapezoid are 628, and 417 
feet. The angles between the longer base and the sides are 
62° 13^, and 84° 38^ Find the lengths of the non-parallel sides. 

4. One side of a parallelogram is 29 feet long, and it makes 
with the diagonals angles of 26° and 44°. Find the length of the 
other side. 

5. The diagonals of a parallelogram are 102 and 76 feet, and 
the angle between them is 58°. Find the lengths of the sides 
and the angles of the figure. 

6. Two circles, whose radii are 31 and 23 inches, intersect, 
the radii drawn to one of the points of intersection making an 
angle of 27° 30^. Find the length of the line joining their centers. 
[Note that the problem has two solutions.] 

7. Three circles whose radii are 12, 16 and 25 inches are tan- 
gent to each other externally. Determine the angles between the 
lines joining their centers. 

8. In a triangle ^ = 46°, & = 121, c = 157. The side a is tri- 
sected and A is joined to the point of trisection nearer to B. Find 
the length of this line. 



76 A SHORT COURSE IN TRIGONOMETRY. 

9. A line through the vertex of an equilateral triangle divides 
the angle into two parts which are in the ratio of 2 : 3. Find the 
ratio in which it divides the opposite side. 

10. To find the width of a river a point B is located on one 
bank, and a line AC^ 550 feet long, is measured on the other bank 
close to the water. The angles ACB and CAB are 65° 2^ and 
56°38^ respectively. How wide is the river? 

11. AB is a straight road running due north from A. At ^ a 
distant point P bears N. 47° E., and at B it bears N. 71° E. If 
AB = 1800 feet, find AP. 

12. A person on the bank of a river observes the elevation of 
the top of a tree on the opposite bank to be 47° 20^. He then 
retires 50 feet horizontally, in a direct line from the tree, and 
observes the elevation to be 44° 35^. How wide is the river? 

13. A and B are two points on a straight horizontal road 2000 
yards long running N. W. and S. E. At ^ a mountain bears N. 
4° E. At B it bears N. 37° E. and has an elevation of 21° 45^ 
Find the height of the mountain above the level of the road, and 
its distance (measured horizontally) from the road. 

14. To find the distance between two inaccessible points, M 
and N, a line AB, 350 feet long is taken, so situated that M and 
N lie on the same side of it. BAM= 102° 19^, BAN =4,1° 7', 
ABN= 98° 15^, ABM= 52° 18^ Determine the length of MN. 

15. To find the distance between two inaccessible points A and 
B, a line CD is measured so that CD cuts AB. CD = 600 feet, 
i>CB = 58°12^, CDB = 49°S7', ACD = 74° 16', ADC=62°13'. 
Find the length of AB. 

16. To find the distance between two inaccessible points, A 
and B, a point C is taken in the line AB, and a point D not 
in that line. CD = 437 feet, i>CB = 114° 42^, ^DC= 65° 7^ 
CDB = 41° 21^ Find the length of AB. 

17. A, B and C are three peaks. From A, B bears N. 65° E., 
and C bears N. 10° W. ; while from B the bearing of C is N. 55° W. 
The elevations of B and C as observed from A are 8° 24^ and 
15° 38^ respectively, and the altitude of B above A is known to 
be 425 feet. Determine the altitude of C above A . 

18. From a point on a beach which slopes uniformly down 
from the base of a cliff" at an angle 6, the elevation of the cliff" 




SOLUTION OF OBLIQUE TKIAXGLES. 77 

is j3. From a point a feet nearer the cliff the elevation is a. De- 
termine the altitude of the cliff above the first point of observa- 
tion. 

Note : The distance a is to be taken as the horizontal distance 
between the two points of observation. 

Solve the problem if ^ = 5°, a = 56°, (3 = 47°, and a =60 feet 

19. An observer h feet above the surface of a lake observes that 
the angle of elevation of a point on the opposite shore is /5, and 
the angle of depression of the image of the same point in the 
water is a. Determine the altitude of the point observed above 
the surface of the lake. 

Solve the problem if /i = 150 feet, a = 25° and 13 = 7°. 

20. From a building h feet high the angles of depression of the 
top and bottom of another building are j3 and a respectively. 
Determine the altitude of the second building. 

Solve the problem if /i = 110 feet, a = 60° and ^ = 20°. 

21. A right triangle rests on its hypotenuse, the length of 
which is 125 feet, with its plane inclined 75° to the horizon. If 
one of its angles is 40°, determine the altitude of its right angle 
from the ground. 

22. A, B and C are three consecutive milestones on a straight 
road, from each of which a distant spire is visible. The spire 
bears N. E. at ^, E. at B and E. 30° S. at C. Find the distance 
of the spire from A, and the shortest distance from the spire to 
the road. 

23. A horizontal line is drawn through the base of a tower, and 
three points A, B, C are taken on this line. At A the angle of 
elevation of the top of the tower is a, at B it is 90° — a and at C 
it is 2a. If AB = a and BC=h, show that the height of the 
tower is V{a + hf — \d\ 

24. At each end of a horizontal base of length 2a, the angle 
of elevation of a peak is /?, and at the middle of the base it is a. 
Prove that the height of the peak above the plane of the base is 

a sin a sin 3 



l/sin (a -f /3) sin (a — /3) 

25. Two circles of radii R and r are in the same plane, and 
the length of the line joining their centers is a. i? > r, and 



78 A SHORT COURSE IN TRIGONOMETRY. 

R-\- r <Ca. Their internal common tangent meets the line of 
centers at an angle a, and their external common tangent meets 
the same line at an angle ^i. Show that 

-R = a sin \{a^[i) cos J(a — ^), r = a cos J(a + /?) sin J(a — /3). 

26. Prove the following relations between the parts of a right 
triangle. 

(1) sin 2 A = sin 2B. (5) cos 2 A = sin {B — A). 

(2) a? cot A = ¥ cot B. (6) cos 2A = (^ + ^) (^ - ^ ), 

(3) tan I A = ^. (7) cos {A -B) = ^-^. 

(4)tan(45o^^)^^. (8) t.HA-B)=^^±^^. 

Prove the following relations between the part of any triangle. 

27. a = b cos C -^ c cos B, 
b = c cos ^ + a cos C, 
c = a cos ^ -f 6 cos A, 

28. a + & + c = (5 -h c) cos ^ + (c + a) cos jB + (« + ^) cos C. 

29. he cos J[ -h ca cos 5 + a6 cos C= |(a^ + 6"^ + c^). 

30. a (6 cos C — c cos 5) = 52 _ ^2^ 

31. a (cos B cos C+ cos A) = h (cos J. cos C+ cos B) 

= c (cos A cos 5 + cos C). 

32. 6 cos -B + c cos C = a cos (5 — C) 

33. 1 — tan 1^ tan J5 = -. 

34. a cos A-^h cos 5 + c cos C = 2a sin 5 sin 0. 

35. a? — 2ab cos (60° + C) = c'' — 2bc cos (60° + A). 



1 



CHAPTEE YI 



MISCELLANEOUS PROBLEMS. 

45. Area of a Triangle. Let K represent the area of 
any triangle, and jp the altitude, CP. Then 

but since jp = h sin A = a sin B, we 
have 




K = Ibc sin A = Jc« sin ^, 1 ,^^^ 

K can also be expressed in terms of the sides of the tri- 
angle alone. If in the first of (67) we substitute 

sin ^ = 2 sin ^A cos J J,, 



= 2 



Vs {s — a){s — b) {s — c) 



by (63) and (64), 
— (68) 



we have K = Vs (« — a) (s — b)(s — c). 

46. Radius of the Inscribed Cir- 
cle.* Let the required radius be 
denoted by r. ^ 

Let AE=AF=x, 

BF= BD = y, 

CD=. CE=z; 

2{x + y -{- z) = a + b -i- G = 2sy 

or X -{- y -]- z ^ s. 

* The circle inscribed in a triangle is sometimes called the incircle. 

79 




80 




A 


SHORT COURSE IN TRIGONOMETRY. 




Then 








X 

y 

z 


= s — {y -{- z) = s — a, 
==, s — (z + x) = s — b, 
= s — (x -^ y) = s — c. 


.^ 


2 

i 


E 


-y 


1. 


jC :n'ow 

r = AE tanj A = x tan J J., 
hence 

r = {s — a) tan J^, 

and similarly 








B 




r = {s — h) tan JJB, 






Fig. 39. 




r = [s— c) tan JC. 





(69) 



If for tan ^ A wq substitute its value taken from (65) 



^= V 



{s — a) {s — h)(s — c) 




(70) 



From the values of tan \A, tan ^B, tan J (7 in (65), 

. -r. , ^ 1 I (s — a) (s — 6) (s — G) r 
tan I A tan J^ tan JC = - a| ^ "^ ^^ ^ = -; 

r = s tan i^ tan ^B tan JC (71) 

47. Radii of the Ex-circles. If the sides of a triangle 
be produced, three circles can be drawn, each of which touches 
one side of the triangle exter- M 

nally, and the other two sides pro- 
duced. These circles are called a- 
escribed circles, or excircles. One 

of the excircles is shown in Fig. ^ N 

40. Its center lies at the inter- 




FiG. 40. 



section of the bisectors of the exterior angles at B and C, 
and the bisector of the angle A. 

From the figure AM = AC -{-CL = b -\-CL, 
AN=AB+ BL=e-{- BL, 



MISCELLANEOUS PROBLEMS. 



81 



AM-\- AN=b + G + (BL + CL), 
= a -{- b -\- c = 2s, 
But AM= ANy hence AM = s. 

Then if the radius of this excircle be represented by r^, 
we have r^ = s tan J^, 



and similarly 



r2 = s tan Jjg, 

r^ = s tan JO, 



(72) 



where 7-^ and rg are the radii of the excircles opposite the 
angles B and C respectively. 

If the values of tan J J., tan J5, tan J (7 from (65) be 
substituted in (72), we have 



^3= ^ 



s{s 


-6)(s- 


-^•) 




s — a 


J 


s(s 


-e){s- 


a) 




s — b 


? 


s{s 


-a){s- 


«') 



(73) 



48. Radius of the Circumscribed Circle.* Let the 

C radius be represented by R. The angle 

BOC= 2A. Hence a = 2B sin A. 

a 




iJ = 



2 sin A 
b c 



(74) 



Fig. 41. 
From (74) ^ = 2 



2 sin ^ 2 sin C J 

a -\- b -^ G 



(sin ^ + sin jB 4- sin C)' 



* The circle circumscribed about a triangle is sometimes called the cir- 
cumcircle. 



82 A SHORT COURSE IN TRIGONOMETRY. 

Now 

sin A + sin jB -f- sin (7 = sin (^ + C) + sin B -f sin (7, 

= 2 sin i(^ + C) cos i{B + C) 

+ 2 sin 1(5 4- C) cos J(J5 - C), 
= 2 sin i(^ + (7) [cos 1(5 H- C) 4- cos J(5 - O)] , 
= 2sini(5+ (7) [2 cos 15 cos JC], * 

= 4 cos J A cos J5 cos J C. 

s 
~ 4 cos ^A cos ^B cos JC7 * ^ "^ 

In (75) substitute for cos J Aj cos J 5, cos J O their 
values from (64), then after reduction, we find 



Exercises. 



1. Showthat- + - + i = - 

^1 ^2 ^3 ^ 

From (73) 



1,11_ Vs — a Vs — h , Vs — c 



n ^2 ^3 /s (s — ^»)(s — c) i/s(.«? — c)(s— a) i/s(s— a)(s-6) 

_ (g — g) + (g — 6) + (s — c) _ 3g — (g + ^> + c ) 

Vs (s — g) (s — 6) (s — c) t/s(s — g)(s — 6)(g — c) 
s l/g 1 



l/s(s — g)(s— 6)(s — c) /(s — g)(s — 5)(s — c) »* 

2. Given E = 100, g = 36, 6 = 71. Solve the triangle. From 
.^.. . . g . _, 6 , ^, gsinC 6 sin C 
(74) sm4 = 2:^' ^'"-^ = 25' ^""^ ^'^^^ " = 1!^ '^'•'' = ^hrB- 
The logarithmic work is left for the student to supply. 

3. r=62, (7= 121° 13^, g=193. Solve the triangle. First 

T 

we have s — c= —^ , see (69). Then since s — h = a — (s — c), 

tan jtG 



miscella:n^eous problems. 83 



we have tan ^B = j-. This gives us all the angles. To 

s — 

T 

complete the solution find s — a by using s — a = ~ ; then, 

Tan 2"^ 

since a is given, we find s by adding s — a and a. Finally h and 

c are found by subtracting s — h and s — c from s. For a check 

use (60). The student may supply the logarithmic work. 

Examples. 

I. Determine K^ r, r^, r^, r^ and B for each of the triangles in 
the examples at the end of § 43. 

2. Prove geometrically that K=rs. 

3. Prove the following relations : 

(1) r^r/g = sK= s'^r. (2) rr^r^r^ = K^. 

(3) r = 4:R sin ^A sin ^B sin ^C. 

(4) r^ = 4E sin ^A cos ^B cos JC. 

(5) r^ = 4:B cos |J. sin ^B cos JC 

(6) r^ = 4jB cos J^ cos ^B sin JC. 

, , ^ a^ sin B sin a^ sin B sin C 

4. Show that K = — ^r—. — -. = =— ^ — .p , ^. . 

2 sm ^ 2 sin (5 + (7) 

Solve the following triangles from the parts given. 

5. B = 43.16, A = 76° 12^, 5 = 59° 8^ 

6. r = 12. 19, ^ = 37° 58^ 5 = 98° 13^ 

7. -K = 100, A = 47°, c = 63. 

8. r = 29.3, 5 = 58°, a = 120. 

9. K= 5729, a = 62, ^ = 73° 12^ 

10. E:= 2126, ^ = 52°, B = 68°. 

II. Construct a triangle ^5C, and let P be the center of the 
inscribed circle, and P^P^P^ the centers of the three ex-circles op- 
posite to ^, B, O respectively. Show 

_csin^5 2a sin. ^Bsm^C 
^ ^ cos JO sin A ' 

with similar expressions for PB and PC. 



84 A SHORT COURSE IN TRIGONOMETRY. 

a COS ^C 2b sin ^A cos I C 



(2) P,B 



RC 



cos ^A sin B ' 

a cos ^B 2c sin |^^ cos ^B 



^ cos J^ sin C * 

with similar expressions for P^C, F^A, P^A, P^B. 
(3) APP, = K-^ + B), APP, = ^{A + C), etc. 

C4^ P P = ^ P P = ^ P P = ^— 



cosj^' 2 cosiP' 3 cos^C" 

„ . 2acos APcos iC 

(6) P^ = 1^—. ~, etc. 

^ ^ ^ sm^ ' 

(7) Area of P,P,P,=^ 2 sin i^ sin JPsin ^0^ 2r ' 
where Z" is the area of ABC. 

12. ^PC is any triangle and M is the middle point of AB ; AD 
is drawn perpendicular to PC, cutting CM" in L ; show that 

. ^ a6 sin (7 

^^= — r^: >f 

a + cos C 

13. If the side BC of a triangle be bisected at D and AD be 
drawn, show that tan ADB = 



b'-c' 



n 



CHAPTER VII. 

SPHERICAL RIGHT TRIANGLES. 

49. Introduction. Spherical trigonometry treats of the 
solution of spherical triangles. 

A spherical triangle is a portion of the surface of a sphere 
bounded by three arcs of great circles of the sphere. From. 
geometry we learn that a spherical triangle is determined 
completely when three parts are given, not excepting the 
case when the given parts are the three angles. 

Since the sum of the angles of a spherical triangle is 
greater than two, and less than six right angles, it is possible 
to have spherical triangles with one, two, or three right angles. 
It is also possible that one, two, or three sides may be quad- 
rants. Hence we have the following definitions : 

I. Spherical triangles having one^ two, or three right angles 
are called respectively spherical right triangles, hirectangular 
triangles, and trirectangular triangles, 

II. If one, two, or three sides of a spherical triangle are 
quadrants, the triangle is called respectively a quadrantal, 
biquadrantal, or triquadrantal triangle. 

It is shown in geometry that a trirectangular triangle is 
also triquadrantal, and is equal to one eighth of the surface 
of the sphere. 

The six elements of a spherical triangle correspond, each 
to each, to the six elements of the trihedral angle which the 
triangle subtends at the center of the sphere. The sides of 
the triangle correspond to the face angles of the trihedral 
angle, and the angles of the triangle to the dihedral angles. 

85 



86 



A SHORT COURSE IN TRIGONOMETRY. 




Fig. 42. 



A figure will make this clear. Let ABC he any triangle 
upon the surface of the sphere whose center is at 0. Let 
the planes of the sides of the triangle be constructed, meet- 
ing, two and two, in the lines OA, 
OB and OC. These three planes 
will form a trihedral angle at 0. 
The face angles of this trihedral an- 
gle, A OB J BOO and CO A are meas- 
ured by c, a and b respectively, and 
the angles of the triangle J., B and 
C have the same magnitude as the 
dihedral angles whose edges are OA, OB and 0(7 respectively. 
In all that follows each of the sides and angles of the 
spherical triangle will be taken less than 180°. 

50. Formulae for the Solution of Spherical Right Tri- 
angles. Let ABC he a spherical right triangle on a sphere 
whose center is O. Through J.', any point on OA pass a 
plane perpendicular to OA, cutting 
the planes A OB, BOC, CO A in 
A'L, LM, A'M respectively. This 
plane is perpendicular to the plane 
AOCy since it is perpendicular to 
OA, Hence LM, being the inter- 
section of two planes both perpen- 
dicular to the plane AOC, is itself 
perpendicular to the plane AOC, 
and is therefore perpendicular to both A' M and OC. More- 
over A'L and A' M are both perpendicular to OA, hence 
LA' Mis the plane angle of the dihedral angle BO AC, and 
therefore the angle LA' M equals the angle A of the spher- 
ical triangle. 

In the right triangle OLM, we have 

ML_ML A'L 

OL ~ A'L ' OL ' 




Fig. 43. 



sm LOM=^^ = 



SPHEEICAL EIGHT TRIANGLES. 87 

LOM=a, -TT-p = sin J.'= sin J., -^r^ = sin J.' OZ = sin c, 

hence sin « = sin ^ sin c. (77) 

Hence, also sin 6 = sin ^ sin c. (78) 

Again in the right triangle OLA' , we have 

OA' OA' OM 



cos A' 0L = 



OL~OM OL' 



OA 



but A' OL = c, yym = ^^s ^' OM— cos 6, 

and -^^y^ = cos MOL = cos a, 

hence cos c = cos a cos &. (79) 

Now using the right triangle A'L3I, we have 
A'3I A'M OA' 



cos LA' M= 



A'L ~ OA' A'L 



A'M 
but LA'M = A, -y^-rr = tan A' OM = tan 6, 

' OA' ' 



and 



-JTT = cot J.' Oi = cot C, 



hence cos A = tan & cot c. (80) 

Hence also cos B = tan a cot c. (81) 

^. cose cos a COS 6 _ _^. 

bmce cotc = -i — = - -. , irom (79), 

smc smc ^ ^ 

we may substitute this value of cot c in (80), and obtain 

cos a cos b cos a sin 6 
cos A = tan 6 



sin c sm c 



But from (78) -^ = sin B, 

^ ^ sm c 

6 



88 A SHORT COURSE IN TRIGONOMETRY. 

hence cos A = cos a sin B^ (82) 

similarly cos ^ = cos 6 sin ^. (83) 

The product of (78) and (81) gives 

sin b cos B = tan a sin B cos c, 

or, since cos c = cos a cos b, 

sin 6 cos ^ = sin a sin B cos b, 

hence, dividing by sin B cos 6, 

sin a = tan b cot ^, (84) 

similarly sin b = tan a cot A. (85) 

Finally, from (82) and (83) 

cos A , cos B 

cos a = —. — 7s» cos o = ^ 7, 

sm B sin J.' 

or cos a cos 6 = cot A cot 5, 

.-. by (79) cos c = cot ^ cot B, (86) 

By means of these ten formulae the unknown parts of any 
spherical right triangle can be computed, when two parts, 
other than the right angle, are given. 

51. Napier's Rules. The ten formulae just derived may 
be arranged in two groups, thus : 

sin a = sin A sin c, sin a = tan b cot B, 

sin b = sin B sin c, sin b = tan a cot A, 

cos c = cos a cos 6, cos c = cot A cot Bj 

cos ^ = sin B cos a, cos ^ = tan b cot c, 

cos B = sin ^ cos b, cos 5 = tan a cot c. 

In all of these formulae write for c, J. and 5 their com- 
plements, expressed as (co. c), (co. A), (co. B), making in 
each case the corresponding change in the function. Then 
we have 




SPHERICAL RIGHT TRIANGLES. 89 

sina = cos (co. A) cos (co. c), sina:=tan b tan (co. JS), 

sin 6 = cos (co. B) cos (co. c), sin 6 = tan a tan (co. A), 

sin (co. c) ^cosa, cos b, sin (co. c) = tan (co. A) tan (co. B), 

sin (co. A) = cos (co. ^) cos a, sin (co. A) = tan 6 tan (co. c), 

sin (co. -B) =cos (co. A) cos b, sin (co. B) =: tan a tan (co. c). 

It now appears that each part or complement of a part of 
the triangle, excepting the right angle, is expressed in the first 
group in terms of the product of two ^ 

cosines, and in the second group in 
terms of the product of two tangents. 
Moreover it will be seen from Fig. 44 
that in the second group the two parts 
on the right of each equation are imme- ^ . . 

diately adjacent to the part on the left, 
while in the formulae of the first group the parts on the 
right are each separated from that on the left by an interven- 
ing part. Thus in the first formula of the first group A is 
separated from a by 6, and c is separated from a by B. 
When the parts on the right of the equation are thus separ- 
ated from the part on the left they are said to be opposite to 
it, otherwise they are said to be adjacent. 
The part on the left of each equation is called 
the middle part. In determming whether 
two parts are opposite or adjacent the right 
angle is entirely ignored. In determining 
whether parts are opposite or adjacent, be- 
ginners may find it convenient to arrange 
them in their proper order around a circle, as shown in Fig. 
45, the right angle being omitted. 

Napier's rules are simply a condensed form of statement 
for these ten formulse. They are 

I. The sine of the middle part equals the product of the co^ 
sines of the opposite parts. 

II. The sine of the middle part equals the product of the tan- 
gents of the adjacent parts. 




90 A SHORT COURSE IN TRIGONOMETRY. 

It will assist the memory to note that the vowel o occurs 
in cosine and opposite, while a occurs in tangent and adjacent. 

52. Theorems. I. In any spherical right triangle a side 
and its opposite angle are hoth'^ 90°, or both < 90°. 

-r^ /^^v . . cos -S 

From (83) sin J. = p- 

^ ^ cos 6 

Now A is < 180°, hence sin A is always positive. There- 
fore cos B and cos h have always the same sign, which proves 
the proposition. 

II. If the two sides of a spherical right triangle are both 
> 90° or both < 90°, the hypothenuse is < 90°; but if one 
side is > 90° and the other <C 90°, the hypothenuse is > 90°. 

From (79), cos c = cos a cos 6. 

If a and b are both> 90° or both < 90°, cos a and cos 6 
have the same sign. Therefore cos c is positive, and c is 
< 90°. If, however, a is > 90° and b < 90°, or vice versa, 
cos a and cos b have opposite signs. Therefore cos c is nega- 
tive and c is> 90°. 

53. Application of Napier's Rules. 

1. Given a = 42° 19', c = 78° 16'. Draw a figure like 
Fig. 44, and proceed as follows. Take any one of the un- 
known parts in connection with the two given parts and 
pick out that one of the trio under consideration which 
is either opposite to or adjacent to both the other parts. 
In this way consider b in connection with the given parts, 
a and c. We see from Figs. 44 or 45 that a and b are ad- 
jacent to each other, while c is opposite to both of them. 
Hence c is the middle part, and the first rule gives 

sin (co. c) = cos a cos 6, or cos c = cos a cos b : 

, cos c , .. 

cos 6 = . (i) 

cos a 



SPHEEICAL RIGHT TRIANGLES. 91 

In the same way let A, c, and a be considered. Then a 
is the middle part, being opposite to both A and c, and, again, 
by Rule I., we have 

sin a = cos (co. A) cos (co. c), or sin a = sin J. sin c : 

'A ^^^^ /--x 

sm A = —. . (ii) 

smc ^ ^ 

Finally, let B, c, and a be considered. Here B is the 
middle part, being adjacent to both c and a. Hence we 
have, by Rule II., 

sin (co. B) = tan a tan (co. c), or cos B = tan a cot c. {Hi) 

(i)y (ii) and (iii) express the unknown parts 6, A and Bj each 
in terms of the given parts c and a. Special attention is 
called to this feature of the solution. If each unknown part 
is found directly from the given ones, as can always be done, 
no error that may be made in the determination of one part is 
transmitted to another ; an advantage that is obvious. 

In order to check the work, make up a formula between 
the three parts found and see w^hether it is satisfied by the 
results. Here the check formula is between Ay B, and 6, of 
which B is the middle part, and we have 

sin (co. B) = cos b cos (co. A), 
or cos B = cos b sin A. 

The logarithmic work is as follows : 

log cos c = 9. 3082 log sin a = 9. 8282 log tan a = 9. 9593 

log cos a = 9. 8689 log sin c = 9.9908 log cot c = 9.3174 

log cos b = 9.4393 log sin ^ = 9. 8374 log cos 5 = 9. 2767 

. _ f 43° 27^ Check. log cos b = 9.4393 

~tl36°33^ log sin ^ ^ 9.8374 

^ = 79° 6^ log cos 5 = 9.2767 



92 A SHORT COUESE IN TRIGONOMETRY. 

Since A is determined by its sine^ the solution gives two 
values for that part ; but I, § 62, shows that the smaller of 
these values is the correct one. Hence the results are 

6 =74° 2', J.= 43°27^ 5 = 79° 6\ 

If the student will arrange his scheme of work as above 
before looking up any of the logarithms, he will be able by 
two openings of the tables (as the tables are generally ar- 
ranged) to take out all six of the logarithms needed for the 
solution. 

54. Ambiguous Case. When in a spherical right tri- 
angle a side and opposite angle are given there will be 
always two solutions. Construct the right triangle ABC 
so as to contain a given angle, A, and a given side opposite, 

a. Produce the sides AB and 
AC until they meet in A^. A 
lune is thus formed, and the angle 
A' equals A. Hence the right 
triangle A^BC contains also the 
given angle and opposite side. The other pairs of corre- 
sponding sides, and the two values of B are respectively sup- 
plementary. 

In the formulae for solving this case each unknown part 
is given by its sine, hence they give the double solution 
directly. 

Example : Given A = 76° 12^,' a = S7° 36^, solve the triangle. 
Taking b with A and a, we have by rule II. 

sin b = tan (co. A) tan a, or sin b = cot A tan a. (^) 

Taking c with A and a, we have by rule I. 

/ ^x / N • sina . .... 

sm a = cos (co. A) cos (co. c), or sm c = - — r- W 

sin^ \ / 

Taking B with A and a, we have by rule I. 

cos A 

sin (co. A) = cos (co. B) cos a, or sin B = . (Hi) 

^ ■ cos a ^ ' 

The check formula is sin 6 = sin c sin B. 




SPHERICAL RIGHT TRIANGLES. 



93 



Logarithmic work : 

log cot^ = 9.3903 log sin a =9.7854 log cos ^ = 9.3776 

log tan a = 9.8866 log sin ^ = 9.9873 log cos a = 9.8989 

log sin b = 9^2769 log sin c = 9.7981 log sin B = 9.4787 

b^ = 10°54^ b^ = 169° 6' check log sin c = 9.7981 



q = 38°59^ C2 = 141° V 



log sin ^ = 9.4787 



B^ = 17°31^ B^ = 162°29^ log sin b = 9.2768 

The parts designated by the same subscript belong to the same 
triangle. 

55. Spherical Isosceles Triangles. To solve a spherical 
isosceles triangle, divide it into two symmetrical right tri- 
angles by a perpendicular from the vertex to the base. 

In the isosceles triangle ABC, let a = 74° a 

28^ B = 54° 12^. Find A and b or c. Let AD 
be the perpendicular. 

Solution : In the right triangle ABD, one side 
!« = 37°14^, and B = 54°12^ are given. Taking 
the angle ^A with these parts we have by Na- 
pier's rule I. 

cos I A = cos la sin B. 

Taking c with ^a and B, we have by rule II. 

r> ^ n ^ X tan la 

cos B = tan la cot c, or tan c = —. 

' cos B 

To check the work use c, I A and B, then by rule II, 

cos c = cot I A cot B. 
Logarithmic work : 




log COS la -- 
log sin B - 

log cos I A -- 


= 9. 9010 log tan ia = 9. 8808 
- 9. 9091 log cos ^ = 9. 7671 
= 9.8101 log tan c = 0.1137 


log cot J^ = 9.9272 
log cot 5 = 9.8581 
log cose = 9.7853 




i^ = 49°47^ ^ = 99°34^ c 


= 52° 25^ 



56. Quadrantal Triangles. It is proved in geometry 
that in two polar triangles each angle of one is the supple- 
ment of the side of the other of which it is the pole. That 
is if ABC and A^B' C , Fig. 48, are polar triangles, in which 
A is the pole of B' C^ , etc., then 



94 



A SHORT COURSE IX TRIGONOMETRY. 



A-\-a' = 180°, A' -\-a=^ 180°, , 

^ + 6' = 180°, B' ^-h^ 180°, 

0+ c' = 180°, C + c = 180°. 

If c = 90°, then ABC\^ a quad- 
ran tal triangle, and A' B' C is a 
right triangle, with the right angle 
at C , Hence to solve a quadrantal 
triangle in which two parts are given 
in addition to the quadrant, we find 
the corresponding parts in the polar 
triangle and solve that. 

Example.— Given A = 121° 12^, b = 78° 43^, c = 90°, solve the 
triangle. In the polar triangle a^ = 180°— J = 58° 48^, B^ = 180° 
— 6 = 101° 17^ C = 180° — c = 90°. 




By Napier's rules we have 
cos A^ = cos a^ sin B^, cot c^ = 



cos B^ 

tan a^ 



tan b^ = 



sin a^ 



cot J5^ 



and for a check formula cos A'' = tan 6^ cot c\ 
Logarithmic work : 

. log cos a^= 9.7143 log cos -B^=(—) 9.2915 logsina^=: 9.9321 
log sin B'= 9.9915 log tan a'= 0.2178 log cot B'= (— ) 9.3000 
log cos A^= 9.7058 log cot c'= (— ) 9.0737 log tan b'= (— ) 0.6321 
A' = 59° 29^ . •. a = 120° 31^ log cot c'= (— ) 9.0737 

c'= 96° 45^ .-.(7= 83° 15^ log cos ^^= 9^7058 

b' = 103° 8^ .. B= 76° 52^ Check. 

Note : In the solution of this example certain logarithms are 
preceded by the symbol (— ). This sign does not affect the loga- 
rithm, but is used to remind the computer that the number cor- 
responding to the logarithm so marked is negative. In deter- 
mining the sign of the result it must be remembered that every 
logarithm so marked indicates a negative factor, and hence if the 
number of such symbols is even the result is positive, otherwise it 
is negative. Some computers write the letter n after the loga- 
rithm to indicate the same thing. Thus in this example they 
would write log cos -B^= 9.291 5?i. 



SPHERICAL RIGHT TRIANGLES. 95 

Examples. 
Solve each of the following right triangles : 

I. c= 140°, a = 20°. 2. a = 75° 5^, 5 = 35° 30^ 
3. c = 62° 18^, A = 103° 2K 4. 5 = 41° 38^ B = 56° 20^ 
5. a= 112°, A = 100°. 6. a= 35° 26^ b = 113° 36^ 

Solve each of the following isosceles triangles ; b and c always 
designate the equal sides. 

7. A = 68° 34^, B = 62° 12^ 8. a = 121° 26^, c = 78° 32^ 
9. a = 15° 54^, B = 68° 17^ 10. c = 106° 15^, B = 110° 6V. 

II. ^ = 28°38^ a = 20°26^ 

12. A = 47° 16^, p = 100°. \ p represents the perpendicular 

13. a = 76° 40^^ p= 25°. j from A to the base. 

Solve each of the following quadrantal triangles. 

14. c = 90°, a = 47°, B = 68°. 

15. c = 90°, A = 96°, B = 59°. 

16. c = 90°, a = 112° 27^ b = 97° 42^ 

17. c = 90°, 6 = 76° 14^, B = 58° 35^ 

18. Show that the three sides of a spherical right triangle must 
be all less than 90°, or two of them must be greater than 90°. 

19. Show that a side and the hypothenuse are in the same or 
different quadrants, according as the included angle is less than 
or greater than a right angle. 

20. Show that in a spherical right triangle each angle is nearer 
90° than its opposite side. 

21. If ^ = a, show that the remaining parts are all equal to 90°. 

22. If each of the sides of an equilateral spherical triangle is 
represented by a, and each of the angles by A, show that sin ^A 

= . Hence compute the angles if the sides are each 

2 cos ^a 

equal to 65°. 

23. A ship starts due east from a port in 41° N. latitude and 
sails on an arc of a great circle of the earth 38° 37^. What is her 
latitude at the end of the journey, and in what direction is she 
sailing ? 

24. An observer takes the altitudes of two stars, one due north 
and the other due west, and finds them to be 52° 9^ and 36° 37^ 
respectively. Find the angular distance between the stars. 



CHAPTEE VIII. 



FOEMUL^ FOR SPHEEICAL OBLIQUE TRIANGLES. 



57. Theorems. 

I. In any spherical triangle the sines of the sides are propor- 
tional to the sines of the opposite angles. 

Let ABC be any spherical triangle, the center of the 
sphere, and OA, OB, and OC the edges of the dihedral 
angles formed by the planes of the sides of the triangle. 

From B^y any point of OB, drawn 
B^P perpendicular to AOC, and 
through B^F pass planes perpen- 
dicular to OA and 0(7 respectively. 
The first of these will intersect the 
planes AOC and A OB in the lines 
PJ.' and B^A^, both perpendicu- 
lar to OA, and the second will in- 
tersect AOC and BOC in PC and 
B^C^, both perpendicular to OC. 
Hence angle B'A'P=:= A, and angle B'C'P= C. 
In the right triangles A'PB' and C'PB' 

B'P B'P 

sin A' = -jj^f = sin A, sin C = j^, p, = sin C. 

sin^ B'C 
Hence -. — -^ = ., p. . (^) 

sm C A'B' ^ ^ 

In the right triangles A' OB' and COB' 

A'B B'C 

sin A' OB' = ^T^F^ = sin c, sin C OB' = ^^. = sin a. 




OB' 



OB' 



Hence 



sing B'C 
sin c ~ A'B' ' 

96 



(ii) 



FORMULAE FOR SPHERICAL OBLIQUE TRIANGLES. 97 



Therefore, from (^) and (n), 

sin a sin A 



sin c sin C 

The same thing may be proved for each pair of sides and 
angles, hence 

sin a sin h sin c 

sin ^ ~ sin JB ~ sin C' 



or. 



sin a sin ^ = sin h sin A, 
sin & sin C = sin c sin ^, 
sin c sin ^ = sin a sin C. 



(87) 



It should be observed that the demonstration is valid 
whether the foot of the perpendicular B' F is inside the sec- 
tor J. 0(7 or outside of it, and that the theorem is therefore 
perfectly general. This remark applies also to that which 
follows. 

II. In any spherical triangle the cosine of each Me is equal 
to the product of the cosines of the other two sides plus the prod- 
uct of the sines of these sides and the cosine of their imiluded 
angle. 

Fig. 50 is constructed in the 
same manner as Fig. 49, with the 
addition of A^M drawn in the 
plane AOC perpendicular to 
OC, and PiV^ parallel to OC. 

PC' 
OF 



PM MC 



PB 
PM 



7 + 



+ 



PB' 
PN 




PB' ' PB 



but 






A 
Fig. 50. 


oc 





98 



A SHOET COUESE IN TEIGONOMETRY. 



OB' ~ OA' ^ OB' ~ ^^^ ^^^ ^' 




PN 
OW 



PN A'P 



and 
A'B' 



A'P ^ A'B' "" OB' 
= sin PA' N cos A sin c, 
= sin b sin c cos Ay 

since PA'N=AOC = b (sides 
perpendicular each to each). Sub- 
stituting in (i) these values of 

OC OM , PN 



OB' , OB' 



and 



OB' 



we have cos a = cos b cos c -f sin b sin c cos -4. 
Similarly, cos & = cos c cos c* -f sin c sin a cos B^V (88) 
cos c = cos c«^ cos b-\- ^m a sin & cos C, 

These formulae furnish a theoretical solution of a triangle 
when two sides and the included angle are given, or when 
the three sides are given ; but they are not well adapted for 
logarithmic computation. Other formulae more useful in 
practical computation will be derived from them. When, 
however, two sides of a triangle and the included angle are 
given, and the third side only is required, (88) can be used 
by means of the following device. 

In the formula 



cos a = cos h cos c + sin 6 sin c cos A, 
let ^ sin ^ = cos h, 

and h cos d = sin h cos A, 

Substituting these values we have 

cos a = k sin d cos c -\- k cos 6 sin c, 
or, by (38) cos a= k sin (6 + c). 



(i) 
(ii) 



FORMULA FOR SPHERICAL OBLIQUE TRIANGLES. 99 

To determine the auxiliary quantities h and 6 divide (i) 

by {ii), 

. cos h cot h 

Then tan d = -. — 7 r = r? 

sm o cos A cos A 

cos 6 , sin 6 cos J. 

and from {%) or (^^) A; = — — ^ , or 



sin ' cos u 

Example. A = 73° 26', h = 38° 12^, c = 49° 37^ find a. 

log cot 6 = 0.1041 _ ( 77°2r log cos h = 9.8953 

log cos^ = 9.4550 ~t257°21^ log sin ^ = (dt) 9.9893 

log tan 6 = 0.6491 _ j 126°58^ log k = (d=) 9^060 

^ ^ ~ 1 306°58^ log sin (^ + c) = (d=) 9.9025 

a = 49°57^ log cos a = 9.8085 

It should be noted that ^ is not limited to values less than 180° 
as are the parts of the triangle, hence both values of 6 less than 
360° are to be taken, and also the corresponding values of 6 -{- c. 
Hence both k and sin {6 + c) have two values numerically equal 
but with opposite signs, but as the upper signs must be taken to- 
gether and also the lower signs, only one value of a less than 180° 
is obtained. In practice therefore it is sufficient to use one value 
of^. 

Examples. 1. Given B = 61° 16', a = 38°, c = 74°. Find b. 
2. Given C = 102°, a = 85°, b = 63°. Find c. 

58. Formulae Deduced by the Aid of the Polar Tri- 
angle. Attention has been called to the fact that in two 
polar triangles, ABC and A'B'C, 

^ = 180°-a^ ^=180°-6^ (7=180°-c^ 

a = 180°-^^ 6=180°-^', c=180°-(7'. 

If these substitutions be made in any formula of spherical 
triangles, a new formula is derived of the same form as the 
original but having angles where the other had sides and 
vice versa. If this is done in the first of (87) the result is 

sin (180° - A') sin (180° - h') 

LoFC. = sin (180° - B') sin (180° - a% 



100 A SHORT COURSE IN TRIGONOMETRY. 

or sin A^ sin 6' = sin B^ sin a', 

which is what we started with. Hence nothing has been 
gained. 

Treating (88) in the same way, we have 

cos (180° - A') = cos (180° - B') cos (180° - O') 
+ sin (1 80° - B') sin (180° - C) cos (180° - a'), 

or — cos A' = cos B^ cos C — sin B' sin C cos a', 

which is a new formula. Dropping accents and changing 
signs we have 

cos A= — cos JB cos O + sin B sin C cos a, ^ 

cosB = — cos C cos ^ + sin C sin A cos &, ;► (89) 

cos O = — cos ^ cos ^ 4- sin A sin B cos c. I 

J 

Formulae (89) may be used to find the unknown angle, 
when two angles and the included side are given. Thus in 

cos A = — cos B cos C -f sin ^ sin Ccos a 
substitute ksin d == — cos B, k cos ^ = sin ^ cos a. 

r^. n cot B 
Then tan /? = , 

cos a 

cos B sinB cos a 

sm cos d ' 

and cos ^ = ^ sin ^ cos C -\-h cos d sin (7, 

or cos J. = ^ sin (^ + C). 

Examples. 1. Given a = 74:°, B = 103°, C= 61°. Find A. 
2. Given c = 38°, A = 51°, B = 69°. Find C. 

59. The Sines of the Half-angles. 

-^ / s . COS a — COS 6 cos c ,^^, 

From (88) cos A = ^^-^ (90) 

^ ^ sm 6 sin c ^ 

Subtracting both sides from unity 



1 



FORMULiE FOR SPHERICAL OBLIQUE TRIANGLES. 101 

sin 6 sin c + cos h cos c — cos a 



1 — cos A = 



sin b sin c 



^ . o . . cos (h — g) — cos a , , , ^^ -, , . . x 

or, 2 sin^ JJL = ^ . , ^ , by (49) and (41). 

' ^ sm 6 sin c ^ *^ ^ ^ ^ ^ 

In (58) let a = a, and ^ = h — c, then 

cos (6 — c) — COS a = 2 sin J- (a + 6 — c) sin J (a — 6 + c) 
= 2 sin (s — c) sin (s — h), 

where s = i (a + 6 + c), see § 35. 

sin (« — 6) sin (s — c) 



Hence sin^ ^A = 

Also 



^ sin & sin c 

sin (s — c) sin (« — a) 
sin c sin a ^ 

sin {s — a) sin [s — &) 



sin^ 1^ = 



(91) 



sin^ ic = 



sin a sin & 

60. The Cosines of the Half- Angles. Adding unity to 
both members of (90) we have 

cos a + sin 6 sin c — cos h cos c 



1 + cos J. = 



sin h sin c 



or, 



2cosH^ = ^''"-''^(^ + ^) 



, by (50) and (39) 



^"" sin h sin c 

In (58) let a = 6 -f c, and j^ = a, then 

cos a — cos (6 + c) = 2 sin J (6 + c 4- a) sin J (6 + c — a) 
= 2 sin s sin (s — a). 



Hence 


00-2 1 ^ ''"' " *'"' ^" '">' 


^ sin b sin c 


Also 


sin « sin (s — b) 


^ sin c sin a ^ 




sin s sin (s — e) 

cos2 JO = ^ \ ^ ^ 

^ sin «^ sin & 



(92) 



102 



A SHOKT COURSE IN TRIGONOMETRY. 



61. The Tangents of the Half- Angles. Dividing each 
of (91) by the corresponding one of (92), we find 



., , sin (s — b) sin. (s — c) 

tan^ iA = ^ r^^ — ^~^— ^ 

sm s sm (s — a) 



tan^ ^B = 



tan^ 1(7 - . . , , . 

sin s sin (« — c) 

If, as in § 37, we use an auxiliary quantity 

sin (« — a) sin (s — 6) sin (« — c) 



sin 


(« 




'^) 


sin 


(* 


— 


«) 




sin 


s 


sin 


(* 


— 


^) 


» 


sin 


(« 


— 


«) 


sin 


(* 


— 


&) 



(93) 



tan^ IT = 



sin s 



(94) 



we have 



tan^ 1 J. = 



or 



sin (s — a) sin (s — b) sin (s — c) tan^ r 



sin s sin^ (s — a) 

tan r 

tan i^ = ^ — ; . , 

^ sin (s — a) 

tan r 
tan 1^ = ^—7 -, 

^ sm (s — b) 



sin^ (s — a) 



tan 1(7 



tan ^^ 



(95) 



n (s — c)* 
62. The Sines of the Half-Sides. 

_ .^^. cos B cos (7-1- cos A 

l^rom (89) cos a = ^ — ^^— ^ — ^^^ 

^ ^ sm ij sm C 

Subtracting both members from unity, 

sin i? sin (7— cos B cos C— cos A 



(96) 



cos a 



or sin^ Ja = 



sin ^ sin C 

cos (^ -h (7) -f- cos A 



2 sm ^ sm C ^ \ / \ / 



FORMULA FOR SPHERieAL OBLIQUE TRIANGLES. 103 

In (57), let a = 5 + Q /9 = ^, then 

cos (5 + C) 4- COS ^ = 2 cos i(B + C+A) cos i{B + C-A), 

= 2 cos /Scos (/S — J-), 
where 8 = i {A + B + C), 

cos S cos (S — A) 



Also, 



sin- Ja = 
sin^ J& = 



sin^ Jc = — 



sin B sin C 

cos S cos (/S — B) 
sin C sin ^ ' 

cos S cos (>Sf — C) 



(97) 



sin ^ sin -B 

63. The cosines of the half-sides. If both members of 
(96) be added to unity, we have 

sin B sin C + cos B cos C + cos A 



or 



1 + cos a = 



cos^ ha = 



sin ^ sin C 

cos (5 — (7) 4- cos J. 



by (50) and (41). 



2 sin B sin C 
In (57), let a= A, and ^ = B — C, then 

cos J. + COS {B-C)= 2 cos i (^ + 5- C) cos J (^-^+ 0), 
= 2 cos (^ - ^) cos {S - C), 



:2l^_ 



cos^ ha = 



cos (/S — B) cos (>Sf — C) 



Also, 



cos 



2 1 



h = 



cos^ ic = 







sin jB sin C 




cos(^ 


— C)cos(S- 


^) 






sin O sin A 




COS 


(« 


- A) cos (.S — 


^) 



(98) 



sin A sin ^ 



64. The Tangents of the Half-Sides. Dividing each of 
(97) by the corresponding one of (98), we have 
7 



104 



A SHORT COUESE IN TRIGONOMETRY. 



tan^ la = — 
tan^ i& = — 



tan^ Jc = 



cos S cos (S — A) 
cos{S— B)cos{S~ cy 

cos S cos (S — B) 
cos(>S — 0)cos(>Sf— Z)' 

cos S cos (aS — C) 
cos {S — A) cos (^S— B)' 



(99) 



I 



Observe that in each of the equations (97) and (99) the 
right-hand members have the negative sign, while the left- 
hand members are perfect squares. This inconsistency, how- 
ever, is only in appearance. We know from geometry that 
28 > 180° and < 540°: therefore 8> 90° and < 270°. 
Hence cos 8 is always negative, and— cos 8 is always positive. 

If we employ in connection with (99) an auxiliary quantity. 

2 — cos ^ 

^^"^ ^ = cos {S - A) cos {S - B) cos (S - cy ^^^^^ 

— cos >S' cos^ (8 — A) 
men tan ^a - ^^^ ^^ _^^ ^^^ ^^ _ ^^ ^^^ ^^ _ ^y 

= tan^ B cos^ (8 - A), 

tan ^a — tan It cos (>S — Ay - 

Similarly tan Jfe = tan JS cos (^ — ^), ► (101) 

tan Jc = tan M cos (>SI — C). . 

65. Napier's Analogies. The product of the first two 
equations of (93) gives 



tan J J. tan ^B 



sin (s — c) 
sin s 



Regarding this as a proportion, we have, by composition and 
division, 

1 + tan J J. tan ^B sin s + sin (s — c) 



tan J J. tan J^ 



sin s — sin {s — c) 



■1^ 



FOEMUL^ FOE SPHERICAL OBLIQUE TRIANGLES. 105 

1 + tan 1 J. tan 15 cos i (A — B) , ^ . . «^ 

^ut .r-^ — 1-T7 — hb = w ( , p x > by Ex. 4, § 30, 

1 — tan ^A tan ^B cos^ (A + B)' -^ ' ' 

sin s 4- sin (s — c) tan i (2s — c) , ,^^^ 

and • -. . ; ( = p-^^ ^ by (59) 

sm s — sin (s — c) tan fc ^ \ / 



tan J (a 4- 6) 
tan Jc 

cos ^ (A — B) tan i (« + 6) 
cos J (^ + J5) ~ tan J c 



(102) 



which is the first of Napier^s Analogies. 

The quotient of the first of (93) by the second gives 

tan ^A sin (s — b) 
tan JjB ~ sin (s — a) 

tan ^A — tan ^B sin (s — b) — sin (s — a) 
tan J J- + tan J^ ~ sin (s — b) -\- sin (s — a) 

tan IJ. — tan ^B sin i (A — B) , ^ ^ ^ ^^ 

But ; hr-n — 1^ = > wj , p6 by Ex. 3, § 30, 

tan 1 J. 4- tan 15 sm ^ (A -{- B) ^ ' ^ ^ 

, sin (s — 6) — sin (s — a) tan J (a —6) 

and -^^ rf-, — ^-7 ( = , — Tj^ ^ by (59) 

sm (s — 6) + sin (s — a) tan J(2s — a — b) '' ^ ^ 

tan J(a — b) ^ 
~ tan Jc 

sin ^(A — B) tan J (<* — ft) 
sin J (^ -f £) ~ tan \c * 

which is the second of Napier's Analogies. 
The product of the first two of (99) gives * 

_ — cos aS' 
tan la tan i6 = y^ 7^ , 

^ ^ cos (8 — C) ' 

* When the product is formed the factor ( — cos S)^ appears on the right, 
and when the square root is taken the negative sign is retained. 



(103) 



106 A SHORT COURSE IN TRIGONOMETRY. 

1 + tan Ja tan J6 cos (6' — C) — cos 8 
1 — tan Ja tan ^b ~ cos (8 — C) -\- cos 8 ' 

1 + tan Ja tan J6 cos J (a — 6) 
1 — tan Ja tan J6 ~~ cos J (a + 6) ' 

^ cos (/S' — (7) — cos >S ^ . /^ rv ^N . -, ^ ^y Ex. 4, 

and — }^ ^^f- ^= tan J (2/S' - C) tan JC, "^ . ^o ' 

cos {8— C) -i- cos8 ^ ^ "^ ^ ' § 33 



= tani(^ + 5)taniC; 

cos !(<* — &) tan ^(A-\- b) 
cos J (a + b) ~ cot JO ^ 



(104) 



which is the third of Napier's Analogies. 

The quotient of the first of (99) by the second gives 

tan ^a cos (8 — A) 
tan 16 "" cos (8 — B) ' 

tan Ja — tan J6 cos (/S' — J.) — cos (8 — B) 
tan Ja + tan J6 ~~ cos (8 — A) -{- cos (/S' — ^)' 

which reduces, as above, to 

"^ smi{a + b)- cotJC ' ^^"^^ 

which is the fourth of Napier's Analogies. 

(104) and (105) may also be derived from (102) and (103) 

by the method of § 58. The student should derive them in 

this way. 

Examples. 

1. Prove the following relations between the parts of any 
spherical triangle. 

cot ^A cot iB cot JC 



I 



sin (s — a) sin (s — b) sin (s — c) 



(2) 



FORMULA FOE SPHERICAL OBLIQUE TRIANGLES. 107 

COS ^A cos ^B sin s 

sin JO sin c 

cos ^A sin Ji? sin (s — a) 



^ ^ cos JC sin c 

sin J^ sin ^B sin (s — c) 



(4) 
(5) 
(6) 

(7) 



sin JC sin c 

sin Ja sin ^b — cos S 

cos |c sin ' 

sin Ja cos J& cos (S — A) 

sin Jc sin 

cos Ja cos \h cos (^ — C) 



cos Jc sin C 

2. If X> be the middle point of 6, show that 

cos c H- cos a = 2 cos \h cos -BD. 

3. In an equilateral triangle show that 

(1) 2 cos A = l — tan^ Ja. (2) sec ^ = 1 + sec a. 

4. If 6 r= c = 2a, show that esc IA = 4: cos a cos Ja. 

5. If a = Itt, show that cos A + cos 5 cos C = 0. 

6. In any spherical triangle bisect a in D and bisect AD in E. 
Then show that 

cos CE+ cos BE =2 cos ^D cos DjEJ. 

7. If c = Jtt, and X be any point on c, show that 

cos CX= cos a sin ^X+ cos 6 sin BX. 

8. Prove in any spherical triangle : 

(1) cos a = cos b cos c -\- sin a sin b sin C cot A. 

(2) tan J(^ — a) tan i(5 + &) = tan ^(A + a) tan i(5 — b). 
sin 2c cos B — sin 26 cos C 



(3) cot a 



cos 26 — cos 2c 



CHAPTER IX. 

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 

66. General Remarks. — The solution of spherical tri- 
angles will be treated under six cases. Before taking them 
up in detail, the student's attention is directed to the follow- 
ing theorems, of which the first four are well known geo- 
metric truths. They will be of assistance in determining the 
quadrant of any of the required sides or angles, when the 
formulae used leave it in doubt. 

(a) The greater side is opposite the greater angle^ and con- 
versely. 

(6) Each side is less than the sum of the other two. 

(c) a + 6 + c < 360°. (d) A + B-^ C> 180°. 

(e) Each angle is greater than the difference between 180° 
and the sum of the other two angles. 

For ^ + ^+(7>180°, 

^>]8o°-(5 4- o): 

If, however, B -\- (7> 180°, then, since in the polar triangle 

a^ <h' + c', 

we have 180° -A< 180° - B -\- 180° - C, 

or _^<l80°-(5+ C); 

^ > (j5 + C) - 180°. 

(/) A side {or angle) which differs more than 90° than 
another side (or angle) is in the same quadrant as its opposite 
angle {or side). 

108 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 109 

The proof for the side is as follows : 

^ ,^^, , cos a — cos 6 cos c 

From (88), cos A = . , . . 

^ ^ sm sm c 

The denominator of this fraction is always positive, and if a 
differs more than 90° than either b or c, we have, neglecting 
signs, 

cos a > cos b, or > cos c ; 
cos a > cos b cos c. 

Hence the sign of the numerator and therefore that of the 
fraction is the same as that of cos a. Therefore cos A and cos a 
have the same sign, and A and a are in the same quadrant. 
The proof for the angles is derived in the same way from 

cos J- + cos -B cos C . ^. 

cos a = -. — ^f^—. — ^ . see (89 ) 

sm B sm C ^ ' 

This theorem may be stated otherwise thus : In a spherical 
triangle only the side and its opposite angle which are nearest 
in value fo 90° can be in different quadrants. 

67. Case I. Given Two Sides and the Included Angle 

a, by and C. 

From (104) tan J (^ + ^) = ^^^ H^ - ^^ ot i Q 
^ ^ ^ ^ ^ cos J (a 4- 6) ^ ' 

from (105) tan H^ - 5) = ^'/^ig"^] cot J C, 

from (103) tan Jc = ^^^ijj^fj tan § (a - b). 

^, , sin a sin 6 sin c 

Check, 



sin A mv B sin C 
Example : a = 121° 17^, h = 76° 31^, C= 50° 12^ 

\{a + 6) = 98° 54^, J(a — &) = 22° 23^, ^C = 25° 6^ 



110 A SHORT COURSE IN TRIGONOMETRY. 

log COS J (a — 6) = 9.9659 log sin ^ (a — 6) = 9.5807 

colog cos ^{a + 6) = (— ) 0.8105 colog sin ^{a + b) = 0.0053 

log cot JO = 0.3293 log cot ^(7= 0.3293 

log tan i(A-^B) = (— ) 1. 1057 log tan i{A — B) = 9. 9153 

log sin i{A + B) = 9.9987 i{A + B) = 94° 29^ A = 133° 56^ 

cologsinJ(^ — 5) = 0.1969 ^{A — B) = S9° 27', B= 55° 2' 

log tan ^a — 6) = 9.6147 

log tan Jc = 9.8103 Jc = 32°52^, c= 65° 44^ 

Check. 

log sin a = 9.9317 log sin 6 = 9.9878 log sin c = 9.9598 

log sin ^ = 9.8574 log sin 5 = 9.9136 log sin 0= 9.8855 

.0743 .0742 .0743 

Examples. 

1. a= 79° 6', 5 = 112° 13^, C= 86° 14^ 

2. a =121° 12^, b= 76° 37^ C = 147° 23^ 
S. b= 64° 23^ c = 100° 49^ A = 95° 38^ 
4. a= 68° 20^ c= 52° 18^, ^ = 117° 12^ 

68. Case II. Given Two Angles and the Included 
Side, A, B, and c. 

From (102) tan J (a + 6) = ^^^ ! j j ~ f j tan Jc, 

sin - f-4 JB) 

from (103) tan i (a - 6) = ^.^ 1(^4-^) ^"^^ i^' 

from (105) cot JC = ^^ ^ [^ ^ ^j tan J (^ - ^). 



Chech 



sin a sin 6 sin c 



sin J. sin 5 sin C 
The logarithmic work may be arranged as in Case I. 

Examples. 

1. A= 65° 23^, ^ = 101° 7% c = 132°12^ 

2.B= 46° 7^ C= 56° 28^ a = 132°46^ 

3. ^ = 121° 36^ C = 34° 15^, b = 50° 10^ 

4:. A= 56° 16^ 5= 45° 4^, c= 96° 20^ 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. Ill 

69. Case III. Given Two Sides and an Angle Opposite 
One of Them, a, h, and A, 

^ /^^v . -r^ sin 6 sin A 

From (87) sm B = , , 

^ ^ sin a 

from (102) ten Jc = °°^||j^fj ten J (a + 6), 
from (104) cot fC = ^fj^j tan H^ + B), 



Check. 



sm c sm a 
sin (7 ~~ sin J. 



The fact that By the first part found, is determined by its 
sine, and will, therefore, have two values, often leads to two 
solutions to this problem. To determine beforehand, from 
the data, whether to look for one solution or two, use ^ QQ (/). 
From this it follows that : If the side whose opposite angle is 
given differs less from 90° than the other given side^ the angle 
first found must be taken in the same quadrant as its opposite 
side, and there will be but one solution. 

It is also true that : If the side whose opposite angle is given 
differs more from 90° than the other given side, there will always 
be two solutions. For the two values of By lead to two values 
of tan Jc and cot J C, as appears from the formulae, and hence 
to two values of c and C. To complete the demonstration it 
must be shown that neither value of B can lead to negative 
values of either tan Jc or cot J (7, in which case the correspond- 
ing part would be > 180°. To show this we have from (88) 

cos a — cos b cos c 



sin c 



sin 6 cos A 



Then, since by hypothesis a differs more from 90° than 6, 
we have, neglecting signs, cos a > cos b and hence cos a 
> cos b cos c. Therefore the sign of the numerator is the 



112 A SHORT COURSE IN TRIGONOMETRY. 

same as that of cos a, and, sin b being positive, the sign of 
the denominator is the same as that of cos A. But a and A 
are in the same quadrant [%QQ {fT\y hence cos a and cos A 
have the same sign. Therefore sin c is always positive and 
c has two values less than 180°. Hence C has also two 
values. 

Example, a = 139° 55^, h = 61° 37^ A = 150° 17^ 
Since a differs more from 90° than b, there will be two solu- 
tions. 

log sin b = 9.9444 J(« + ^) = 100° 46' 

log sin ^ = 9.6953 Ka — b) = 39° 9' 

colog sin a =J). 1912 ^{A + B;)= 96° 27^7 

1{A-B^)= 53° 49^3 
^{A -f B^) = 143° 49^3 
^{A-B,)= 6° 27^.7 
log cos l{a -^b) = (— ) 9.2714 
cologcosJ(a — b) = 0.1104 
log tan ^{A + B^) = (— ) 0.9459 
logcotjCl= 0.3277 
log cos ^{a -hb) = (— ) 9.2714 
colog cos ^{a — 6) = 0. 1 1 04 
log tan i{A + B^) = (— ) 9.8641 
logcotJC2= 9.2459 
iCi = 25°10^9, ^C,= 80° 0^5, 
q-=50°21^8, C2==160°l^ 

Check. 

log sin Ci = 0.0000 log sin c^ = 9.6473 log sin a = 9.8088 

log sin Oj = 9.8866 log sin C; = 9.5338 log sin A = 9.6953 

Til34 ~Tll35 .1135 

Examples. 
1. a = 99° 40^ b = 64° 23^ A = 95° 38^ 

2.b= 98° 17^ c = 74°37^ C = 61° 13^ 

3. a = 100° 30^, c = 69°30^, ^ = 80° 20". 

4. c = 89° 28^ b = 59° 50^ C = 56° 28^ 



log sin B = 


-- 9.8309 


A = 


-- 42° 38^4 


S.= 


= 137°2r.6 


logcosi(^ + A) = 


:(-) 9.0513 


colog cos ^{A — jBj) = 


0.2289 


logtan^(a + Z>) = 


:(-) 0.7209 


log tan Jci = 


0.0011 


log cos 1{A 4- B^) = 


:(-) 9.9069 


colog COS ^{A — B^ = 


0.0028 


log tan \{a^b) = 


:(-) 0.7209 


log tan |C2 = 


0.6306 


K = 45°4^4, K = 


76° 49^5, 


c^ = 90°8^8, c^ = 


153° 39^ 



SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 113 

70. Case IV. Given Two Angles and the Side opposite 

one of them, A^ By and a. 

^ .r.^. .7 sin 5 sin a 
From (87) sin b = -. — -. — , 

cos ^ (A -\- B) 
from (102) tan Jo = ^^^||^— ^j tan i(« +6), 

from (104) cot iC= Zll^^l^) ^^H^ + S). 



Check. 



sin c sm a 
sin C~ sin J.' 



The logarithmic work may be arranged as in Case III. 

This case presents the same peculiarities as the preceding, 
because a triangle given in this way can be solved by pass- 
ing to the polar triangle, the solution of which will then fall 
under Case III. Hence the first triangle will have one or 
two solutions according to the number which the polar triangle 
has. The rules by which to determine the number of solu- 
tions to expect are obtained from those printed in italics 
under Case III., by interchanging the words side and angle. 







Examples. 




1. 


A= 76° 6^ 


B= 85° 22^ 


b= 93° 18' 


2. 


A = 132° 16^ 


(7=139°44^ 


c = 127° 30' 


3. 


^ = 120° 0^, 


B= 70° 0^ 


6 = 100° 0' 


4. 


A = 128° 19^, 


B= 70° 0^ 


a = 142° 16' 



71. Case V. Given the Three Sides, a, b, and c. 
This case is most conveniently solved by (94) and (95), 
sin (s — a) sin (s — b) sin {s — c) 



tan'^ r = 



sm 



, . tan r 

tan ^A = -. — -. r , etc. 

^ sm (s — a)' 

_ J sin a sin b sin c 



sin A sin B sin C 



114 A SHORT COURSE IN TRIGONOMETRY. 

Example, a = 112° 7^ h = 127° 39^ c = 71° 12^ 

2s = 310° 58^ colog sin 8 = 0.3820 

s = 155°29^ log sin (s — a) = 9.8368 

8 — a= 43° 22^ log sin (s — b) = 9. 6692 

s — 6 = 27° 50^ log sin {s — c) = 9.9978 

s — c= 84° 17^ log tan^ r = 01858 

log tan r = 9. 9429 log tan r = 9. 9429 log tan r = 9. 9429 

log sin {s—a)= 9.8368 log sin {s—b)= 9.6692 log sin {s—c)= 9.9978 

log tan ^A = 0.1061 log tan ^B = 0. 2737 log tan J C = 9.9451 

1^= 51° 56^ ^B= 61° 58^ iC=41°23^ 

^ = 103° 52^ 5 = 123° 56^ (7=82°46^ 

log sin a = 9.9668 log sin b = 9.8986 log sin c = 9.9762 

log sin A = 9. 9871 log sin B=9. 9189 log sin C = 9. 9965 

9.9797 9.9797 9.9797 

Examples. 

1. a= 106° 48^, b = 85° 5^, c = 97° 55^ 

2. a = 123° 19^ 6 = 67° 37^, c = 164° 12^ 

72. Case VI. Given the Three Angles, A, B, and C, 

Use (100) and (101). 

„ _ — cos 8 

tan^ R = 



Check 



cos {8 -A) cos (/S' - J5) cos (^ - C) ' 
tan Ja = tan R cos ('S' — -A), etc. 
sin a sin 6 sin c 



sin J. sin B sin (7* 
The logarithmic work may be arranged as in Case V. 

Examples. 
1. ^ = 106°19^ B= 87° 13^ C= 96° 48^ 

2.^= 64° 19^ B = U2°27', C=100°00^ 

73. Area of a Spherical Triangle. It is shown in geom- 
etry that the area of a spherical triangle is equal to the area 
of a trirectangular triangle multiplied by the excess of the 



SOLUTION OF SPHERICAL OBLIQUE TRIAJN'GLES. 115 

sum of the angles of the given triangle over two right angles, 
the unit of measure for angles being the right angle. That 
is if K is the area of the triangle, T the area of a trirect- 
angular triangle, and 28= A -\- B -{■ (7, 

jr=rx?^^^. (106) 

Now J'= jTTr^, where r is the radius of the sphere. Hence 
if the angles be expressed in radians. 



K= hrtr 



2 Itt ' 

2'^ 



or 

and if r be unity 



K={2S^7r)r', (107) 

K=2S-7T. (108) 



Exercise. Upon a sphere, whose radius = 10 feet, a triangle 
has A = 72° 32^, B = 58° 6^ C= 101° 9'. Find K. 

By (107), 

3107 
K= 100 X lAOAA ^ square feet = 90.38 square feet. 
lOoOU 

Examples. 
Compute the distance in geographical miles between the points 
named in the first three examples, regarding the earth as a sphere. 
One geographical mile equals one minute of arc of a great circle. 

1. New Orleans, lat. 30° N., long. 90° W., and Havana, lat. 
23° 7' N., long. 82° 30^ W. 

2. San Francisco, lat. 37° 48^ N., long. 123° W., and Manila, 
lat. 14° 30^ N., long. 121° E. 

3. Philadelphia, lat. 40° N., long. 75° W., and Melbourne, 
lat. 38°25^S., long. 145° E. 

4. A steamer sails from Valparaiso, lat. 33° S. , long. 71° 30^ W. 
to Shanghai, lat. 31° 12^ N., long. 121° 30^ E. on an arc of a great 
circle. Determine (1) the number of geographical miles traversed. 



116 A SHORT COURSE IN TRIGONOMETRY. 

(2) the course on leaving Valparaiso, (3) the course on arriving at 
Shanghai, (4) the course on crossing the equator. It is assumed 
that the arc joining these points intersects no islands. 

5. Determine the area, on a sphere of unit radius, of the tri- 
angles in Examples 1 and 2, § 72. 

6. A spherical triangle on the surface of the earth has for its 
three angles A = 73° 25^, B =^ 57° 31^, (7= 51° 13^, determine its 
area in square miles. Take the radius of the earth equal to 4000 
miles. 

7. The area of an equilateral triangle is one fourth the area of 
the sphere. What are its sides and angles ? 

8. The face angles at the apex of a triangular pyramid are 
38° 19^, 46° 12^, and 41° 5^ Determine the dihedral angles. 



TABLES. 



EXPLANATION OF THE TABLES. 



1. Scope of the Tables; Rules. These tables consist 
of three parts : (a) a table of logarithms of numbers ; (6) a 
table of the logarithmic sines, cosines, tangents and cotan- 
gents of angles ; (c) a table of natural sines, cosines, tan- 
gents and cotangents. The term natural sine, cosine, etc., is 
used to denote the actual numerical value of the function. 

The logarithms are all given to four places of decimals, 
and are computed to the base 10. 

The theory of logarithms belongs to algebra, and will not 
be discussed here. The following rules governing the use 
of logarithms in computation are repeated for convenience 
of reference : 

1. To multiply numbers, find the logarithm of each factor, 
and add them ; the sum is the logarithm of the product. 

II. To divide one number by another, subtract the logarithm 
of the divisor from the logarithm of the dividend; the differ- 
ence is the logarithm of the quotient. 

III. To raise a number to any power, multiply the loga- 
rithm of the number by the exponent of the power ; the product 
is the logarithm of the required power of the number. 

IV. To extract any root of a number, divide the logarithm 
of the number by the index of the root; the quotient is the 
logarithm of the required root of the number. 

2. Characteristic and Mantissa. A logarithm consists, 
usually, of two parts : a whole number, called the character- 
istic, and an incommensurable decimal fraction, called the 
mantissa. The table gives only the mantissa ; the charac- 
teristic, which may be positive, negative, or zero, must be 

iii 



log 1000 = 


3, 


because 10^ = 1000 


log 100== 


2, 




10^ = 100 


log 10 = 


1, 




10^ = 10 


log 1 = 


0, 




100 = 1 


log .1 = 


-1, 




10-1 _ I 


log .01 = 


-2, 




10-2= .01 


log .001 = 


-3, 




10-3= .001 



IV EXPLANATION OF THE TABLES. 

supplied in every case by the computer. The mantissa is 
always positive, except in the logarithms of exact powers of 
10, when it is zero. 

Since 10 is the base we have : 



(a) 



This series of equations can be extended indefinitely in both 
directions. 

I. The mantissa of the logarithm of a number depends only 
upon the sequence of figures in the number. 

For take any two numbers, say 729.3 and 7.293. Then 
since 729.3 = 100 x 7.293, we have by § 1, I, log 729.3 
= log 100 + log 7.293 = 2 + log 7.293, and it follows that 
the logarithms differ only by the whole number 2. In 
general therefore since a change in the position of the deci- 
mal point is equivalent to multiplying or dividing by a power 
of ten, the effect on the logarithm is merely to alter its value 
by a whole number. 

II. The characteristiG of the logarithm of a number greater 
than unity is one less than the number of significant figures to 
the left of the decimal point. 

For an examination of (a) shows that for all numbers 
greater than 1 and less than 10 (all numbers with one signifi- 
cant figure before the decimal point) the logarithm is greater 
than and less than 1, that is, its characteristic is ; for all 
numbers greater than 10 and less than 100 (all numbers with 
two significant figures before the decimal point) the logarithm 
is greater than 1 and less than 2, that is, its characteristic is 
1 ; and so on. 



I 



EXPLANATION OF THE TABLES. V 

III. The characteristic of the logarithm of a number less 
than unity is negative^ and is numerically one greater than the 
number of ciphers between the decimal point and the first sig- 
nificant figure. 

Again, from (a) it appears that if a number is greater 
than .1 and less than 1, its logarithm is between and — 1 ; 
that is, using a positive mantissa, which we always do, it is 
— 1 + the mantissa, hence the characteristic is — 1. If the 
number is greater than .01 and less than .1, the logarithm is 
between — 1 and — 2, which is written — 2 + the mantissa ; 
that is, the characteristic is — 2 ; and so on. 

As negative characteristics are awkward to handle in 
practice, it is customary to increase all negative character- 
istics by 10. The beginner will do well to write — 10 after 
the logarithm in such cases, but he will find that he can soon 
discontinue this practice without risk of error. In place of 
III, we have, therefore, the following working rule : 

IV. The characteristic of the logarithm of a number less 
than unity is found by subtracting from 9 the number of 
ciphers between the decimal point and the first significant 
figure. 

Verify the following statements : 

characteristic of log 763.92 = 2 



a 



log 1.9841 = 

log .07296 = — 2, or 8 

log 26 = 1 

log 400000 = 5 

log .9426 = — 1, or 9 

log 3869= 3 

log .00042= —4, or 6 

3. To Find the Logarithm of a Given Number. Look 
in Table I in the column marked iV, on the left of the page, 
for the first two significant figures in the given number, then 
follow this line across the page until you reach the column 



VI EXPLANATION OF THE TABLES. 

which is marked at the top with the third significant figure. 
If the given number has only three significant figures the 
number you find at that point in the table is the mantissa 
of the required logarithm. Thus to find log 486, find 48 in 
the N column and go across the page to the column marked G. 
The number found in the table is 6866, which is the mantissa 
of log 486. Since by § 2, II, the characteristic is 2, we have 
log 486 = 2.6866. 

If the given number has less than three figures annex 
ciphers on the right until it has three figures and proceed as 
before. Thus to find log 7, look for log 7.00 which is found 
to be .8451. 

If the given number has four significant figures take 
the first three figures and proceed as before, and then apply 
a correction for the fourth figure. The procedure is best 
shown by an example. Let it be required to find log 32.87. 
Neglecting the characteristic the table gives log 32.8 = .5159, 
and log 32.9 = .5172. The diiference between these two 
logarithms, which is 13, is called the tabular difference. As 
will be seen the tabular difference varies in diiferent parts of 
the table. It is large at the beginning and small at the end. 
A tabular difference 13 means that the mantissa changes by 
13 for a change of 1 unit in the third place of the number. 
The corresponding change for 7 in the fourth place is found 
by multiplying the tabular diiference by .7, and similarly 
in other cases. Thus here the correction is 13 x. 7 = 9.1, 
the nearest integer to which is added to log 32.8 to obtain 
log 32.87. Hence, log 32.87 = 1.5159 + .0009 = 1.5168. 

If the given number has more than four figures make a 
correction for the fifth figure one tenth as large as would be 
made for the same figure in the fourth place, and so on. Ex- 
cept near the beginning of the table the correction thus found 
for figures after the fourth will be too small to affect the 
value of the logarithm. The process of finding the loga- 



EXPLANATION OF THE TABLES. VU 

rithm of a number whose value is between two of the num- 
bers given in the table is called interpolation. 
Verify the following statements : 

log 863.2 =2.9361, log 3 =0.4771, 

log .04312 -=8.6347 (-10), log .0796 == 8.9009 ( — 10), 

log 1.29 =0.1106, log .00183 = 7. 2625 ( — 10), 

log 18000 = 4.2553, log 4762 = 3.6778, 

log .89 =9. 9494 ( — 10), log 123.46 = 2.0915. 

4. To Find the Number which Corresponds to a Given 
Logarithm. Given log x= 1.5786 to find x. Disregard 
the characteristic and look in the body of the table for the 
given mantissa. In this case we find 5786 in the same line 
with the number 37 in the N column, hence these are the 
first two figures of the required number. The third figure 
is the one found at the top of the column which contains the 
given mantissa, in this case 9. Since the given characteristic 
is 1, there must be two figures before the decimal point (§2, 
II.). Hence the required number is 37.9. 

In most cases the exact mantissa will not be found in 
the table, and a correction must then be applied. Suppose 
log X = 0.4223. The nearest mantissse in the table are 4216 
and 4232. The first corresponds to 264 and the second to 
265. The value oi x is therefore between these, and can be 
found by interpolation. Subtract from the given mantissa 
(4223) the one next smaller (4216). The difference is 7, 
while the tabular difference is 4232-4216 = 16. The 
question then is to determine what change in the number cor- 
responds to a change of 7 in the logarithm when a change of 
16 in the logarithm corresponds to a change of 1 in the third 
place of the number. This is determined by the proportion 
16 : 7 = 1 : required correction ; that is, required correction 
= ■j'g = .4. Hence the required number x is greater than 
264 by .4 of a unit in the third place. That is a; = 2.644, 
since the characteristic of log x is 0. 



Vlll EXPLANATION OF THE TABLES. 

Above, iu reducing the fraction -^^ to a decimal, and in all 
similar cases, we keep only the nearest number of tenths. 
This is a special application of two practical rules which 
must always be observed. 

I. Do not carry the work beyond the number of decimal 
places given in the table, that is with this table, four places. 

II. When the division is not exact, always take in the last 
place the figure that is nearest the true result. 

Verify the following statements : 

logx= 1.7368, X = 54.55, log re = 9.7446 — 10, a; = .5554, 

log re = 5.4161, X = 260700, logx = 4.4849, x = 30540, 

log re = 8. 9192 — 10, re =.08302, log re = 3.1413, re = 1385, 

log re = 2.4362, re = 273.0, log re = 7.7901 — 10, re = .006167, 

log re = .6444, re = 4.410, log re = 6.5682 — 10, re =.00037. 

5. The Arithmetical Complement of the Logarithm 
or Co-logarithm. To compute the value of a/b by loga- 
rithms, we may take either log a — log b, or log a + log 1/6. 
Log 1/6 = log 1 — log 6 = — log 6 is called the co-logarithm 
of 6. We have, therefore, the following rule : 

To form the co-logarithm of a given number subtract the 
logarithm of the number from 0. 

It is customary in practice to subtract the logarithm from 
10 instead of from 0, and then to write — 10 after the re- 
sult ; that is, the logarithm is subtracted from 0, written in 
the form 10.00000 — 10. If the logarithm is one which 
has been itself augmented by 10, the two — lO's, the one in 
the subtrahend and the one in the minuend, cancel each other. 
When the characteristic of the logarithm is 0, care must be 
taken not to forget to subtract this from 9, just as any other 
characteristic would be subtracted. 

The practical advantage of using cologs consists in the fact 
that thereby the number of separate operations required to 
obtain the log of the result is reduced. For example, suppose 



EXPLANATION OF THE TABLES. IX 

we wish to calculate log -^ >. Without using cologs 

three operations are required : 

(1) to find log a + log h + log c, 

(2) " logd + loge+log/, 

(3) to subtract (2) from (1). 

If, on the other hand, cologs are used, these three opera- 
tions are reduced to one, viz. : to find log a + log 6 + log c 
-f colog d 4- colog e + colog /. 

Ex. Find colog 729.6. Log 729.6 = 2.8631. Subtracting this 
from 10.00000 — 10, the result is colog 729.6 = 7.1369 — 10. 

Ex. Find colog .0641. Log .0641 = 8.8069 —10. Subtracting 
this from 10.00000 — 10, the result is colog .0641 = 1.1931. 
Verify the following statements : 

colog 9986 = 6.0006 — 10, colog 3.9 = 9.4089 — 10, 
colog 7. 298 = 9. 1368 — 10, colog 380. 6 = 7 .4195 — 10, 
colog .4682 = .3296, colog .005 = 2.3010. 

Exercises. 1. Compute the value of (1.789)^ 

By III. § 1, we have log (1.789)^ = 5 X log 1.789. 
log 1.789 = .2526 

5 

log (1.789)5= 1.2630 .-. (1.789)5 = 18.32. 

2. Compute the value of 728 X 63.86 X -4792 

log 728 = 2.862-1 
log 63.86 = 1.8052 
log .4792 = 9.6805 — 10 

.-. by L § 1, log (728 X 63.86 X .4792) = { ^^ ^^'g^^g ~ ^^ 

Hence 728 X 63.86 X .4792 = 22270. 

3. Compute the value of #'73. 

^ log 73 = 1.8633. 
By IV, § 1, log f 73 = 1 log 73 = .6211, 
f73= 4.179. 

4. Find V^. 06398. 

log.06398 = 8.8061 — 10. 



X EXPLANATION OF THE TABLES. 

We cannot divide this logarithm by 7 without getting an awk- 
ward result. But if we add and subtract 60, we have 

log .06398 = 68.8061 — 70, 

where the number subtracted from the logarithm is now ten time's 
the number by which we must divide ; and hence, after the divi- 
sion, it will be reduced to 10. This is the best practice for such 
cases. Performing the division, we have 



n 



log v^ .06398 -=9.8294 — 10, .-. y .06398 = .6752. 

r; ^27 . , 
o, x = ; , find X. 

(9261)^' 

log l/27 = i log 27 =iX 1.4314 = .7157 

log (9.261)^ = f log 9.261 = f X 0.9667 = .4143 

By 11. §1 log a; =.3014 

a; = 2.002. 

68.96 X 1^74228 _ . 
Q, x = 5 — , find X. 

39X(8.642;3 X (.96)2' 

log 6 8.96 = 1.8386 

log ^.4228 = 9.8754 
colog 39 = 8.4089 

colog (8.642)§ = 8.4390 
colog (.96)2 =0.0354 

logo: =8T5973 .-. a; = .03956. 

EXAMPLES. 

Find the values of the following numerical expressions, and 
give the results to four significant figures : 

1. 839.6X/6I29. ^718.65730. 5 21. 38 X 6.296 X .412 ^^^ ^293 

' 7 X f 41290 

2. 19.63Xr689.2.^ns.l73.4. ^^^ ^^ ^-^^ 

'' (3.339)3X142.9 •-^--- -001984. 

'' ^x (2:^r ^- '■'''- ,.298^gxv:n^ ^^^ ^^^^ 

f 12 X V^n 1^41.63 X (2.649)^ 



EXPLANATION OF THE TABLES. XI 

6. The Table of Logarithmic Sines, Cosines, Tangents 
and Cotangents. This table gives the logarithm of the sin, 
cos, tan and cot of angles from 0° to 90°. From 0° to 5° 
and from 85° to 90° the functions are given for every min- 
ute, and throughout the rest of table for every 10 minutes. 
The general arrangement of the table should be studied first. 
For angles less than 45° the table reads downward, the 
angle being on the left, and the function being read at the 
top of the page ; while for angles greater than 45° the table 
is read upward, the angle being on the right and the function 
being read at the bottom of the page. 

In those parts of the table where the functions are given 
for every 10 minutes the corrections for intermediate num- 
bers of minutes are computed in the same manner as in 
Table I. Since the sine and tangent of an angle increase 
as the angle increases, while the cosine and cotangent de- 
crease as the angle increases, we have the rule : 

When the function is sin or tan add the correctionj but when 
it is cos or cot subtract the correction. 

In general with four-place logarithms angles are computed 
to the nearest minute, but in parts of the table where the dif 
ferences are large (15 or more for 10' intervals) fractions of 
a minute may be used. 

It should be noted that since the sin and cos of all angles 
(except sin 90° and cos 0°), and the tan of angles less than 
45° are proper fractions, the corresponding logarithms have 
negative characteristics. These logarithms all appear in the 
tables with characteristics increased by 10. 

If log sec or log esc is desired find log cos or log sin respec- 
tively, and take the colog. For since sec 6 = ^, log sec d 

= colog cos d, etc. 



Xii EXPLAJIATION OF THE TABLES. 

Exercises. 1. Find log siu 24° 38^. 

The table gives log sin 24° 30' = 9.6177 

Tabular difference = 28 

Correction for 8' = 28 X .8 = 22 

. *. log sin 24° 38' = 9. 61 99 

2. Find log cos 57° 44'. 

From the tables log cos 57° 40' = 9.7282 

Tabular difference = 20 

Correction for 4' = 20 X .4 = 8 

.-. log cos 57° 44' = 9.7274 

3. Find log tan 74° 26'. 

From the tables log tan 74° 20' = 0.5521 

Tabular difference = 49 

Correction for 6' = 49 X -6 = 29 

. •• log tan 74° 26' = 0.5550 

4. Find log cot 37° 3'. 

From the tables log cot 37° 0' = 0.1229 

Tabular difference = 26 

Correction for 3' = 26 X . 3 = 8 

. •. log cot 37° 3' = 0.1221 
Verify the following statements : 

log sin 39° 16' = 9.8014, log cos 8° 19' = 9.9954, 

log tan 63° 24' = 0.3004, log cot 54^ 9' = 9.8583, 

log cos 41° 31' = 9.8744, log tan 82° 56' = 0.9067, 

log cot 26° 12' = 0.3080, log cot 7° 2' = 0.9088, 

log cos 31° 5' = 9.9327, log sin 19° 12' = 9.5170. 

7. To Find an Angle, Given One of its Logarithmic 
Functions. If the given logarithm is log sin or log cos 
look for it in the first or fourth column of logarithms in the 
table, if it is log tan or log cot look in the second or third 
column. If the exact logarithm is found in the tables, or if 
not, when you have located the nearest logarithm that is 
smaller, see whether the given function is named at the top 
or bottom of the page. If at the top read the angle on the 
left, if at the bottom read the angle on the right. When 



EXPLANATION OF THE TABLES. XIU 

the exact logarithm is not found determine the correction to 
be applied in the same manner as in Table I. If the func- 
tion is sin or tan the correction is to be added, but if it is 
cos or cot the correction is to be subtracted. 

Exercises. 1. Log sin 6 = 9.8659, what is ^ ? 

The nearest log sin less than the given one is 9.8653, which 
corresponds to 47° 10^. The tabular difference is 12, and the dif- 
ference between the given log sin and log sin 47° 10^ is 6. Hence 
the correction to be added is x^2 of 10^ = 5^- 

.'.6 = 47° 15^, and 180° — 47° 15^ = 132° 45^ 

2. Log cot d = 0.7265, what is ^? 

The nearest log cot less than 0.7265 is 0.7250, and the corre- 
sponding angle is 10° 40^. The tabular difference is 70, and 
0.7265 — 0.7250 = 15. Hence the correction to be subtracted 
is ^ of 10^ = 2\ 

.-. ^ = 10° 38^ and 180° + 10° 38^ = 190° 38^ 

3. Log tan d = 8.6125, what is ^ ? 

This log tan falls in that part of the table where the functions 
are given for every minute. It is between log tan 2° 20^ and log 
tan 2° 21^, but is nearer the latter. Hence 6 = 2° 2V, and 180° 
+ 2° 21^ = 182° 21^ 

4. Log cos d={—)^ 9.6298, what is ^? 

Find first the angle where log cos is 9.6298. The nearest log 
cos in the table is 9.6286 which corresponds to 64° 50^. The 
tabular difference is 27, and 9.6298 — 9.6286 = 12. Hence the 
correction to be subtracted is |f of 10^ = 4^. Therefore the angle 
is 64° 46^ But cos 6 is negative, hence 6 = 180° =b 64° 46^ 
= 115° 14^ and 244° 46^ 

Verify the following statements : 



log sin e = 




9.2864, 


6= 11° 9^ and 168° 51^ 




log cos 6 = 




8.4632, 


6= 88° 20^ '' 271° 40^ 




log tan 6 = 




0.1298, 


e= 53° 26^ " 233° 26^ 




log cot e = 




9.6241, 


ft= 67° 11^ '' 247° IV 




log sin 6 = 


(- 


-) 9.9641, 


6 = 247° 2' " 292° 58^ 




log cos 6 = 


(- 


-) 9.7841, 


^=127° 28^ '' 232° 32^ 




log tan 6 = 


(- 


-) 9.4231, 


^ = 165° 10^ '' 345° 10^ 




log cot d = 


(- 


-) 8.7643, 


6= 93° 20^ " 273° 20^ 



* This sign ( — ) indicates that cos is a negative number. It does not 
affect the logarithm. 



XIV 



EXPLANATION OF THE TABLES. 



EXAMPLES. 

Find e in each of the following examples : 
6.298 sin2 63° 18^ 



1. 



tan = 



cos d 



7.596 cot 116° 36^ 

2.93 tan 48° 7^ 
14.12 sin 26° 14^ 



sin e 



cot 



-4 



sin^ 146° 12^ X tan 78° 13^ 
cot^ 12° 14^ X cos 64° 4^ 



.9386 cos2 312 ° 39^ 
.8647 tans 214"°~27^ 



e=' 


[127° 0^ 
[307° 0^ 


6= < 


[121° 35^ 

238° 25^ 


6= ■ 


■ 7° 58^ 
172° 2' 
187° 58^ 
352° 2^ 


( 32° 56^ 
~ (212° 56^ 



Note. — The student may vary his practice in the matter of 
interpolation. Thus, to find log 4.287, instead of looking for log 
4.28 and adding the correction for 7, he may look for log 4.29 
and subtract the correction for 3. To find log sin 27° 39^ he may 
look for log sin 27° 40^ and subtract the correction for V instead 
of using the process described above. Also, in finding an angle 
from log cos or log cot it will often be convenient to take the next 
larger log in the tables and add the correction, instead of taking 
the next smaller log and subtracting the correction. Thus, to 
find e from log cot 6 = .3876, we note that 67° 50^ has 3900 for its 
log cot. The difference is 24, and tabular difference 64. Here 
we add |f of 10^ = 4% and 6=67° 54^ 

8. The Table of Natural Functions. This table gives 
the actual numerical values of the sin, cos, tan and cot of 
angles from 0° to 90° at intervals of 10 minutes. The 
arrangement is identical with that of Table II. and the rules 
of interpolation the same. 



TABLE I, 



LOGARITHMS OP JSTUMBERS- 



jr. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


lO 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


II 

12 
13 
14 
15 


0414 
0792 

1139 
1461 
1761 


0453 
0828 

1173 
1492 
1790 


0492 
0864 
1206 
1523 
1818 


0531 
0899 
1239 

1553 
1847 


0569 

0934 
1271 

1584 
1875 


0607 
0969 

1303 
1614 
1903 


0645 
1004 

1335 
1644 

1931 


0682 
1038 
1367 
1673 
1959 


0719 
1072 

1399 
1703 
1987 


0755 
1 106 
1430 
1732 
2014 


16 

17 
18 

19 
20 . 


2041 
2304 

2553 
2788 
3010 


2068 
2330 
2577 
2810 

3032 


2095 
2355 
2601 

2833 
3054 


2122 
2380 
2625 
2856 
3075 


2148 
2405 
2648 
2878 
3096 


2175 
2430 
2672 
2900 
3118 


2201 

2455 
2695 

2923 
3139 


2227 
2480 
2718 

2945 
3160 


2253 
2504 
2742 
2967 
3181 


2279 

2529 
2765 

2989 
3201 


21 
22 

23 
24 
25 


3222 

3424 
3617 
3802 
3979 


3243 
3444 
3636 
3820 

3997 


3263 
3464 

Itli 

4014 


3284 
3483 
3674 
3856 
4031 


3304 
3502 
3692 

3874 
4048 


3324 
3522 
3711 
3892 
4065 


3345 
3541 
3729 
3909 

4082 


3365 
3560 
3747 
3927 
4099 


3385 
3579 
3766 

3945 
4116 


3404 
3598 
3784 
3962 

4133 


26 

27 
28 
29 
30 


4150 
4314 
4472 
4624 
4771 


4166 
4330 

4487 
4639 
4786 


4183 
4346 
4502 

4654 
4800 


4200 
4362 
4518 
4669 
4814 


4216 

4378 
4533 
4683 
4829 


4232 
4393 
4548 
4698 

4843 


4249 
4409 

4564 
4713 
4857 


4265 
4425 
4579 
4728 
4871 


4281 
4440 
4594 
4742 
4886 


4298 
4456 
4609 

4757 
4900 


31 
32 

33 
34 
35 


4914 
5051 
5185 
5315 
5441 


4928 
5065 
5198 
5328 
5453 


4942 

5079 
5211 

5340 

5465 


4955 
5092 
5224 
5353 
5478 


4969 
5105 
5237 
5366 
5490 


4983 
5119 
5250 
5378 
5502 

5623 
5740 
5855 
5966 
6075 


4997 
5132 
5263 
5391 
5514 


5011 

5145 
5276 

5403 
5527 


5024 

5159 
5289 
5416 
5539 


5038 
5172 
5302 
5428 
5551 


36 

37 
38 

39 
40 


5563 
5682 

5798 
591 1 
6021 


5575 
5694 
5809 
5922 
6031 


5587 
5705 
5821 

5933 
6042 


5599 

5717 
5832 
5944 
6053 


5611 
5729 
5843 
5955 
6064 


5635 

5752 
5866 

5977 
6085 


5647 
5763 
5877 
5988 
6096 


5658 

5775 
5888 

5999 
6107 


5670 
5786 

5899 
6010 
6117 


41 
42 

43 
44 
45 


6128 
6232 
6335 
6435 
6532 


6138 
6243 
6345 
6444 
6542 


6149 
6253 
6355 
6454 
6551 


6160 
6263 

6365 
6464 
6561 


6170 

6274 

6375 
6474 

6571 


6180 
6284 

6385 
6484 
6580 


6191 
6294 
6395 
6493 
6590 


6201 
6304 
6405 
6503 
6599 


6212 
6314 
6415 
6513 
6609 


6222 

6325 
6425 
6522 
6618 


46 

:^ 

49 
50 


6628 
6721 
6812 
6902 
6990 


6637 
6730 
6821 
691 1 
6998 


6646 

6739 
6830 
6920 
7007 


6656 
6749 
6839 
6928 
7016 


6665 

till 

6937 
7024 


6675 
6767 
6857 
6946 

7033 


6684 
6776 
6866 

6955 
7042 


6693 
6785 
6875 
6964 
7050 


6702 
6794 
6884 
6972 
7059 


6712 
6803 

6981 
7067 


51 

52 
53 
54 
55 


7076 
7160 
7243 
7324 
7404 


7084 
7168 
7251 
7332 
7412 


7093 
7177 
7259 
7340 
7419 


7101 
7185 
7267 
7348 
7427 


7110 
7193 
7275 
7356 

7435 


7118 
7202 
7284 
7364 
7443 


7126 
7210 
7292 
7372 
7451 


7135 
7218 
7300 
7380 
7459 


7143 
7226 
7308 
7388 
7466 


7152 
7235 
7316 
7396 

7474 


N. 





1 


2 


3 


4 


5 


6 


1 


8 


9 



Jf. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 
57 
58 

59 
60 


7482 
7559 
7634 
7709 
7782 


7490 
7566 
7642 

7716 
7789 


7497 

7574 
7649 

7723 
7796 


7505 
7582 
7657 
7731 
7803 


7513 
7589 
7664 
7738 
7810 


7520 

7597 
7672 

7745 
7818 


7528 
7604 
7679 
7752 
7825 


7536 
7612 
7686 
7760 
7832 


7543 
7619 
7694 
7767 
7839 


7551 
7627 
7701 

7774 
7846 


61 
62 

63 

64 
65 


7853 
7924 

7993 
8062 
8129 


7860 

7931 
8000 
8069 
8136 


7868 
7938 
8007 

8075 
8142 


7875 
7945 
8014 
8082 
8149 


7882 
7952 
8021 
8089 
8156 


7889 
7959 
8028 
8096 
8162 


7896 
7966 

8035 
8102 
8169 


7903 
7973 
8041 
8109 
8176 


7910 

7980 
8048 
8116 
8182 


7917 
7987 
8055 
8122 
8189 


66 

67 
68 
69 

70 


8195 
8261 

8325 
8388 

8451 


8202 
8267 
8331 
8395 
8457 


8209 
8274 
8338 
8401 
8463 


8215 
8280 

8344 
8407 
8470 


8222 
8287 

8351 
8414 
8476 


8228 
8293 

8357 
8420 
8482 


•8235 
8299 
8363 
8426 
8488 


8241 
8306 
8370 
8432 
8494 


8248 
8312 
8376 

8439 
8500 


8254 
8319 
8382 

8445 
8506 


71 
72 

73 
74 
75 


8513 
8573 
8633 
8692 

8751 


8519 
8579 
8639 
8698 

8756 


8525 
8585 
8645 
8704 
8762 


8531 
8591 
8651 
8710 
8768 


8537 
8597 
8657 
8716 

8774 


8543 
8603 
8663 
8722 
8779 


8549 
8609 
8669 
8727 
8785 


8555 
8615 
8675 
8733 
8791 


8561 
8621 
8681 
8739 
8797 


8567 
8627 
8686 

8745 
8802 


76 

77 
78 

79 
80 


8808 
8865 
8921 
8976 
9031 


8814 
8871 
8927 
8982 
9036 


8820 
8876 
8932 

8987 
9042 


8825 
8882 
8938 
8993 
9047 


'8831 
8887 

9053 


8837 
8893 
8949 
9004 
9058 


8842 
8899 

8954 
9009 
9063 


8848 
8904 
8960 

9015 
9069 


8854 
8910 
8965 
9020 
9074 


8859 
8915 
8971 
9025 
9079 


81 
82 
83 
84 
85 


9085 
9138 
9191 

9243 
9294 


9090 

9143 
9196 
9248 
9299 


9096 

9149 
9201 

9253 
9304 


9101 

9154 
9206 
9258 
9309 


9106 

9159 
9212 
9263 
9315 


9112 
9165 
9217 
9269 
9320 


9117 
9170 
9222 
9274 
9325 


9122 

9175 
9227 

9279 
9330 


9128 
9180 
9232 
9284 
9335 


9133 
9186 
9238 
9289 
9340 


86 

87 
88 
89 
90 


9345 
9395 
9445 
9494 
9542 


9350 
9400 
9450 
9499 
9547 


9355 
9405 
9455 
9504 
9552 


9360 
9410 
9460 
9509 
9557 


9365 
9415 
9465 
9513 
9562 


9370 
9420 
9469 
9518 
9566 


9375 
9425 
9474 
9523 
9571 


9380 

9430 

9479 
9528 

9576 


9385 
9435 
9484 
9533 
9581 


9390 

9440 

9489 
9538 
9586 


91 
92 

93 
94 
95 


9590 
9638 
9685 
9731 

9777 


9595 
9643 
9689 

9736 

9782 


9600 
9647 
9694 
9741 
9786 


9605 
9652 
9699 

9745 
9791 


9609 
9657 
9703 
9750 

9795 


9614 
9661 
9708 

9754 
9800 


9619 
9666 
9713 
9759 
9805 


9624 
9671 
9717 

9763 
9809 


9628 

9675 
9722 
9768 
9814 


9633 
9680 

9727 
9773 
9818 


96 

97 

98 

99 
100 


9912 
9956 
0000 


9827 
9872 
9917 
9961 
0004 


9832 
9877 
9921 

9965 
0009 


9836 
9881 
9926 
9969 
0013 


9841 
9886 
9930 
9974 
0017 


9845 
9890 

9934 
9978 
0022 


9850 
9894 
9939 
9983 
0026 


9854 
9899 
9943 
9987 
0030 


9859 
9903 
9948 
9991 

0035 


9863 
9908 
9952 
9996 
0039 


If. 





1 


2 


3 


4 


5 


6 


7 


8 


» 



Z^^mm7 



1 



TABLE II. 

LOGARITHMIC SINES, COSINES, TANGENTS 
AND COTANGENTS. 







0° 










0° 






/ 


L. Sin. 


L.Tan. 


L. Cot. L.Cos. 




r 


L. Sin. 


L. Tan. 


L. Cot. 


L. Cos. 


30 


o 








0.0000 


60 


30 


7.9408 


7.9409 2.0591 


0.0000 


I 


6.4637 


6.4637 


3.5363 


0.0000 


59 


31 


7.9551 


7.9551 


2.0449 


0.0000 


29 


2 


6.7648 


6.7648 


3.2352 


0.0000 


58 


32 


7.9689 


7.9689 


2.03 1 1 


0.0000 


28 


3 


6.9408 


6.9408 


3.0592 


0.0000 


57 


33 


7.9822 


7.9823 


2.0177 


0.0000 


27 


4 


7.0658 


7.0658 


2.9342 


0.0000 


56 


34 


7.9952 


7.9952 


2.0048 


0.0000 


26 


5 


7.1627 


7.1627 


2.8373 


0.0000 


55 


35 


8.0078 


8.0078 


1.9922 


0.0000 


25 


6 


7.2419 


7.2419 


2.7581 


0.0000 


54 


36 


8.0200 


8.0200 


1.9800 


0.0000 


24 


7 


7.3088 


7.3088 


2.6912 


0.0000 


53 


37 


8.0319 


8.0319 


1. 9681 


0.0000 


23 


8 


7.3668 


7.3668 


2.6332 


0.0000 


52 


38 


8.0435 


8.0435 


1.9565 


0.0000 


22 


9 


7.4180 


7.4180 


2.5820 


0.0000 


51 


39 


8.0548 


8.0548 


1.9452 


0.0000 


21 
20 


lO 


7.4637 


7.4637 


2.5363 


0.0000 


50 


40 


8.0658 


8.0658 


1.9342 


0.0000 


II 


7.5051 


7.5051 


2.4949 


0.0000 


49 


41 


8.0765 


8.0765 


1.9235 


0.0000 


19 


12 


7.5429 


7.5429 


2.4571 


0.0000 


48 


42 


8.0870 


8.0870 


1.9130 


0.0000 


18 


13 


7.5777 


7.5777 


2.4223 


0.0000 


47 


43 


8.0972 


8.0972 


1.9028 


0.0000 


17 


.14 


7.6099 


7.6099 


2.3901 


0.0000 


46 


44 


8.1072 


8.1072 


1.8928 


0.0000 


16 


15 


7.6398 


7.6398 


2.3602 


0.0000 


45 


45 


8.1169 


8.1170 


1.8830 


0.0000 


15 


i6 


7.6678 


7.6678 


2.3322 


0.0000 


44 


46 


8.1265 


8.1265 


1.8735 


0.0000 


14 


17 


7.6942 


7.6942 


2.3058 


0.0000 


43 


47 


8.1358 


8.1359 


1. 8641 


0.0000 


13 


i8 


7.7190 


7.7190 


2.2810 


0.0000 


42 


48 


8.1450 


8.1450 


1.8550 


0.0000 


12 


19 


7.7425 


7.7425 


2.2575 


0.0000 


41 


49 


8.1539 


8.1540 


1.8460 


0.0000 


II 


20 


7.7648 


7.7648 


2.2352 


0.0000 


40 


50 


8.1627 


8.1627 


1.8373 


0.0000 


10 


21 


7.7859 


7.7860 


2.2140 


0.0000 


39 


51 


8.1713 


8.1713 


1.8287 


0.0000 


9 


22 


7.8061 


7.8062 


2.1938 


0.0000 


38 


52 


8.1797 


8.1798 


1.8202 


0.0000 


8 


23 


7.8255 


7.8255 


2.1745 


0.0000 


37 


53 


8.1880 


8.1880 


1.8120 


9.9999 


7 


24 


7.8439 


7.8439 


2.1561 


0.0000 


36 


54 


8.1961 


8.1962 


1.8038 


9.9999 


6 


25 


7.8617 


7.8617 


2.1383 


0.0000 


35 


55 


8.2041 


8.2041 


1.7959 


9.9999 


5 


26 


7.8787 


7.8787 


2.1213 


0.0000 


34 


56 


8.2119 


8.2120 


1.7880 


9.9999 


4 


27 


7.8951 


7.8951 


2.1049 


0.0000 


33 


57 


8.2196 


8.2196 


1.7804 


9.9999 


3 


28 


7.9109 


7.9109 


2.0891 


0.0000 


32 


58 


8.2271 


8.2272 


1.7728 


9.9999 


2 


29 


7.9261 


7.9261 


2.0739 


0.0000 


31 


59 


8.2346 


8.2346 


1.7654 


9.9999 


I 


30 


7.9408 


7.9409 


2.0591 


0.0000 


30 


60 


8.2419 


8.2419 


1.7581 


9.9999 







L.Cos. 


L.Cot. L.Tan. 


L. Sin. 


' 




L. Cos. 


L. Cot. 


L.Tan. 


L. Sin. 


f 



89° 



89' 







1* 


3 










1 


9 






/ 


L. Sin. L. Tan. L. Cot. i L. Cos. 




/ 


L. Sin. L. Tan. L. Cot. L. Cos. 


30 


o 


8.2419 8.2419 1.7581 19-9999 


60 


30 


8.4179 8.4181 1.5819 9.9999 


I 


8.2490J 8.249111.75091 9.9999 


59 


31 


8.4227 


8.4229I 1.5771 


9.9998 


29 


2 


8.256118.256211.7438:9.9999 


58 


32 


8.4275 


8.4276 1.5724 


9.9998 


28 


3 


8.2630! 8.2631 


1.7369 9.9999 


57 


33 


8.4322 


8.4323 


1-5677 


9-9998 


27 


4 
5 


8.2699] 8.2700 


1.7300 9.9999 


56 


34 


8.4368 


8.4370 


1.5630 


9-9998 


26 
25 


8.2766 


8.2767 


1.7233 


9-9999 


55 


35 


8.4414 


8.4416 


1-5584 9.9998 


6 


8.2832 


8.2833 


I.7167 


9.9999 


54 


36 


8.4459 


8.4461 


1-5539 9-9998 


24 


7 


8.2898 


8.2899 


I.7IOI 


9.9999 


53 


37 


8.4504 


8.4506 


1-5494 


9.9998 


23 


8 


8.2962 


8.2963 


1.7037 


9-9999 


52 


38 


8.4549 


8.4551 


1-5449 


9-9998 


22 


9 

lO 


8.3025 
8.3088 


8.3026 
8.3089 


1.6974 
I.69II 


9-9999 


51 


39 


8.4593 


8.4595 


1-5405 


9.9998 


21 
20 


9-9999 


50 


40 


8.4637 


8.4638 


1-5362 


9-9998 


II 


8.3150 


8.3150 


1.6850 


9-9999 


49 


41 


8.4680 


8.4682 


I-5318 


9.9998 


19 


12 


8.3210 8.3211 


1.6789 


9-9999 


48 


42 


8.4723 


8.4725 


1-5275 


9-9998 


18 


13 


8.3270 8.3271 


1.6729 9.9999 


47 


43 


8.4765 


8.4767 


1-5233 


9.9998 


17 


14 


8.3329 8.3330 1.6670I 9.9999 


46 


44 


8.4807 


8.4809 1.5 191 


9-9998 


16 
15 


15 


8.3388 


8.3389! 1. 661 1 j 9.9999 


45 


45 


8.4848 1 8.4851! 1.5 149 


9.9998 


i6 


8.3445 


8.3446 


1.6554 9.9999 


44 


46 


8.4890! 8.4892! 1.5 108 9.9998 


14 


17 


8.3502 


8.3503 


1. 64971 9-9999 


43 


47 


8.4930, 8.4933, 1.5067 9.9998 


13 


i8 


8.3558 


8.3559 


1.644119-9999 


42 


48 


8.4971; 8.4973; 1.5027 9.9998 


12 


19 


8.3613! 8.3614 


1.6386 9.9999 


41 


49 


8.501 1 ! 8.5013^ 1.4987 9.9998 


II 
10 


20 


8.3668, 8.3669 


I-6331 


9-9999 


40 


50 


8.5050: 8.5053! 1.4947' 9-9998 


21 


8.3722! 8.3723 


1.6277 


9-9999 


39 


51 


8.5090, 8.5092 1.4908 9.9998 


9 


22 


8.3775i 8.3776 


1.6224 9.9999 


38 


52 


8.51291 8.5i3r 1.4869 9.9998 


8 


23 


8.3828: 8.3829 


1.6171 9.9999 


37 


53 


8.5167; 8.5170, 1.4830I 9.9998 


7 


24 


8.3880I 8.3881 


1.6119,9.9999 


36 


54 


8.5206 8.5208 1.4792I 9.9998 


6 
5 


25 


8.393118.3932 


1.6068; 9.9999 


35 


55 


8.5243, 8.5246; 1.4754 9.9998 


26 


8.3982; 8.3983 


1.6017 


9-9999 


34 


56 


8.5281 8.5283 1.4717 9-9998 


4 


27 


8.4032 


8.4033 


1-5967 


9.9999 


33 


57 


8.5318 8.5321 1.4679 9.9997 


3 


28 


8.4082 


8.4083 


I-5917 


9-9999 


32 


58 


8.5355 8.5358: 1.4642! 9.9997 


2 


29 


8.4131 


8.4132 


1.5868 


9-9999 


31 


59 


8.5392 8.5394! 1.4606: 9.9997 


I 



30 


8.4179 


8.4181 


1.5819 


9-9999 


30 


60 


8.5428 8.5431 : 1.4569; 9-9997 




L.Cos. 


L.Cot.'L.Tan. 


L. Sin. 


/ 




L. Cos. L. Cot. L. Tan. ! L. Sin. 



88< 



88° 



/ 


L. Sin. L. Tan. 


L.Cot. 


L.Cos. 




' 


L. Sin. L. Tan. L. Cot. L. Cos. 


50 

49 
48 
47 
46 

45 
44 
43 
42 
41 
40 




I 

2 

3 

4 


8.5428 

8.5464 
8.5500 

8.5535 
8.5571 


8.5431 
8.5467 
8.5503 
8.5538 
8.5573 


1-4569 
1-4533 
1.4497 
i.z^462 
1.4427 


9.9997 
9.9997 
9-9997 
9-9997 
9-9997 


60 

59 
58 
57 
56 


10 
II 
12 
13 
14 


8.5776 
8.5809 
8.5842 
8.5875 
8.5907 


8.5779 
8.5812 

8.5845 
8.5878 
8.591 1 


1.4221 
1.4188 

I-4155 
1.4122 
1.4089 


9.9997 
9.9997 
9-9997 
9-9997 
9.9997 


5 
6 

7 
8 

9 


8.5605:8.5608 1.4392 
8.5640! 8.5643 1.4357 
8.5674 8.5677 1.4323 
8.5708 8.571 1 1.4289 
8.5742; 8.5745 1.4255 


9-9997 
9-9997 
9-9997 
9.9997 
9.9997 


55 
54 
53 
52 
51 


15 
16 

17 
18 

19 


8.5939 
8.5972 
8.6003 
8.6035 
8.6066 


8.5943 
8.5975 
8.6007 
8.6038 
8.6070 


1.4057 
1.4025 

1.3993 
1.3962 

1.3930 


9-9997 
9.9997 

9-9997 
9-9996 
9.9996 


10 


8.5776:8.5779:1.42219.9997 


50 


20 


8.6097 8.6101 1.3899 9.9996 




L. Cos. L. Cot. L. Tan. 


L. Sin. 


/ 




L. Cos. L. Cot. L. Tan. L. Sin. 



87° 



87< 



XI 







2 









« 




2 









/ 


L. Sin. L. Tan. L. Cot. L. Cos. 




' 


L. Sin. L. Tan. 


L. Cot. 


L.Cos. 




20 


8.6097 8.6101 1.3899 9.9996 


40 


40 


8.6677 8.6682 


I..3318 


9-9995 


20 


21 


8.6128I 8.6132! 1.3868 9.9996 


39 


41 


8.6704 


8.6709 


I.3291 


9-9995 


19 


22 


8.6159 8.61631 1.3837' 9-9996 


38 


42 


8.6731 


8.6736 


1.3264 


9-9995 


18 


23 


8.61S9 


8.6193: 1.3807; 9.9996 


37 


43 


8.6758 


8.6762 


1.3238 


9.9995 


17 


24 


8.6220 


8.6223 


1.3777! 9-9996 


36 


44 


8.6784 


8.6789 


I.3211 


9-9995 


16 


25 


8.6250 


8.6254 


1. 3746 1 9-9996 


35 


45 


8.6810 


8.6815 


1.3185 


9.9995 


15 


26 


8.6279 


8.6283; 1.3717I 9.9996 


34 


46 


8.6837 


8.6842 


I.3158 


9-9995 


14 


27 


8.6309 


8.6313 


1. 36871 9.9996 


33 


47 


8.6863 


8.6868 


I.3132 


9-9995 


13 


28 


8.6339 


8.6343 


1-3657I 9-9996 


32 


48 


8.6889 


8.6894 


I.3106 


9-9995 


12 


29 


8.6368 


8.6372 


1.3628; 9.9996 


31 


49 


8.6914 


8.6920 


1.3080 


9.9995 


II 


30 


8.6397 


8.6401 


1-3599 


9.9996 


30 


50 


8.6940 


8.6945 


1.3055 


9-9995 


10 


31 


8.6426 


8.6430 


1-3570 


9.9996 


29 


51 


8.6965 


8.6971 


1.3029 


9.9995 


9 


32 


8.6454 


8.6459 


I-3541 


9.9996 


28 


52 


8.6991 


8.6996 


1.3004 


9-9995 


8 


33 


8.6483 


8.6487 


I-3513 


9.9996 


27 


53 


8.7016 


8.7021 


1.2979 


9.9994 


7 


34 


8.65 1 1 


8.6515 


1.3485 


9.9996 


26 


54 


8.7041 


8.7046 


1.2954 


9-9994 


6 


35 


8.6539 


8.6544 


1.3456 


9.9996 


25 


55 


8.7066 


8.7071 


1.2929 


9.9994 


5 


36 


8.6567 


8.6571 


1.3429 


9.9996 


24 


56 


8.7090 


8.7096 


1.2904 


9.9994 


4 


37 


8.6595 


8.6599 


1. 3401 


9-9995 


23 


57 


8.7115 


8.7121 


1.2879 


9-9994 


3 


3« 


8.6622 


8.6627 


1-3373 


9.9995 


22 


58 


8.7140 


8.7145 


1.2855 


9.9994 


2 


39 


8.6650 


8.6654 


1.3346 


9-9995 


21 


59 


8.7164 


8.7170 


1.2830 


9.9994 


I 


40 


8.6677 


8.6682 


1.3318 


9.9995 


20 


60 


8.7188 


8.7194 


1.2806 


9.9994 







L. Cos. 


L. Cot. 


L. Tan. 


L. Sin. 


' 




L.Cos. 


L. Cot. 


L.Tan. 


L. Sin. 


/ 



87' 



87^ 







3 













3 









/ 


L. Sin. 


L.Tan. 


L. Cot. 


L.Cos. 




r 


L. Sin. 


L.Tan. 


L. Cot. L.Cos. 







8.7188 


8.7194 


1.2806 


9.9994 


60 


20 


8.7645 


8.7652 


1.2348 9.9993 


40 


I 


8.7212 


8.7218 


1.2782 


9.9994 


59 


21 


8.7667 


8.7674 


1.2326 


9.9993 


39 


2 


8.7236 


8.7242 


1.2758 


9.9994 


58 


22 


8.7688 


8.7696 


1.2304 


9.9992 


38 


3 


8.7260 


8.7266 


1.2734 


9.9994 


57 


23 


8.7710 


8.7717 


1.2283 


9.9992 


37 


4 


8.7283 


8.7290 


1.2710J 9.9994 


56 


24 


8.7731 


8.7739 


1. 2261 


9.9992 


36 


5 


8.7307 


8.7313 


1.2687 


9.9994 


55 


25 


8.7752 


8.7760 


1.2240 


9.9992 


35 


6 


8.7330 


8.7337 


1.2663 


9-9994 


54 


26 


8.7773 


8.7781 


1.2219 


9.9992 


34 


7 


8.7354 


8.7360 


1.2640 


9-9994 


53 


27 


8.7794 


8.7802 


1. 2198 


9.9992 


33 


8 


8.7377 


8.7383 


1. 2617 


9.9994 


52 


28 


8.7815 


8.7823 


1.2177 


9.9992 


32 


9 


8.7400 


8.7406 


1-2594 


9.9993 


51 


29 


8.7836 


8.7844 


1.2156 


9.9992 


31 


10 


8.7423 


8.7429 


1.2571 


9.9993 


50 


30 


8.7857 


8.7865 


1.2135 


9.9992 


30 


II 


8.7445 


8.7452 


1.2548! 9.9993 


49 


31 


8.7877 


8.7886 


1.2114 


9.9992 


29 


12 


8.7468 


8.7475 


1.25251 9.9993 


48 


32 


8.7898 


8.7906 


1.2094 


9.9992 


28 


13 


8.7491 


8.7497 


1.2503 9.9993 


47 


33 


8.7918 


8.7927 


1.2073 


9.9992 


27 


14 


8.7513 


8.7520 


1.2480 9.9993 


46 


34 


8.7939 


8.7947 


1.2053 


9.9992 


26 


15 


8.7535 


8.7542 


1.2458 9.9993 


45 


35 


8.7959 


8.7967 


1.2033 


9.9992 


25 


16 


8.7557 


8.7565 


1.2435! 9.9993 


44 


36 


8.7979 


8.7988 


1. 2012 


9.9991 


24 


17 


8.7580 


8.7587 


1.2413 


9-9993 


43 


37 


8.7999 


8.8008 


1. 1992 


9.9991 


23 


18 


8.7602 


8.7609 


1.2391 


9-9993 


42 


38 


8.8019 


8.8028 


1. 1972 


9.9991 


22 


19 


87623 


8.7631 


1.2369 


9-9993 


41 


39 


8.8039 


8.8048 


1.1952 


9.9991 


21 


20 


8.7645 


8.7652 


1.2348 


9-9993 


40 


40 


8.8059 


8.8067 


1. 1933 


9.9991 


20 




L. Cos.! L. Cot. L.Tan. 


L. Sin. 


/ 




L. Cos. 


L. Cot. 


L.Tan. 


L. Sin. 


/ 



86° 



86^ 



' 1 L. Sin. 


L.Tan.j L. Cot. L. Cos. 




t 


L. Sin. L.Tan. 


L. Cot. L. Cos. 




40 


8.8059 


8.8067! 1. 1933 9.9991 


20 


50 


8.8251' 8.8261 


I-I739 9-9990 


10 


41 


8.8078 


8.8087 


1.1913 


9.9991 


19 


51 


8.8270 8.8280 


1. 1720 9.9990 


9 


42 


8.8098 


8.8107 


1.1893 


9.9991 


18 


52 


8.8289I 8.8299 


1.1701 9.9990 


8 


43 


8.8117 


8.8126 


1. 1874 


9.9991 


17 


53 


8.8307 8.8317 


1. 1683 9.9990 


7 


44 


8.8137 


8.8146 


1. 1854 


9.9991 


16 


54 


8.8326 8.8336 


1. 1664 9.9990 


6 


45 


8.8156 


8.8165 


1-1835! 9-9991 


15 


55 


8.8345! 8.8355 


1. 1 645 9-9990 


5 


46 


8.8175 


8.8185 


1.1815 


9.9991 


14 


56 


8.8363 


8.8373 


1. 1 627 9.9990 


4 


47 


8.8194 


8.8204 


1. 1796 


9.9991 


13 


57 


8.8381 


8.8392 


1. 1 608 9.9990 


3 


48 


8.8213 


8.8223 


1. 1777 


9.9990 


12 


58 


8.8400 


8.8410 


1-1590 9.9990 


2 


49 


8.8232 8.8242 


1-1758 


9.9990 


II 


59 


8.8418J 8.8428 


1.1572 


9.9989 


I 


50 


8.8251 


8.8261 


1.1739 


9.9990 


10 


60 


8.8436! 8.8446 


i.i554i 9-9989 







L. Cos. 


L. Cot. L.Tan. 


L. Sin. 


1 




L. Cos. 


L.Cot.jL.Tan.lL.Sin. 


1 



86' 



86^ 







4^ 


3 










4 









1 


L. Sin. 


L.Tan. L. Cot.' L. Cos. 




' 


L. Sin.lL.Tan. 


L.Cot. L.Cos. 







8.8436 


8.8446 1. 1554 9-9989 


60 


30 


8.8946', 8.8960 


1. 1040: 9.9987 


30 


I 


8.8454 


8.84651 1. 1535 9.9989 


59 


31 


8.8962 


8.8976 


1.1024, 9.9986 


29 


2 


8.8472 


8.8483 


1-1517, 9-9989 


58 


32 


8.8978 


8.8992 


1. 1008 


9.9986 


28 


3 


8.8490 


8.8501 


1. 1499 9-9989 


57 


33 


8.8994 


8.9008 


1.0992 


9.9986 


27 


4 


8.8508 


8.8518 


1.1482} 9.9989 


56 


34 


8.9010 


8.9024 


1.0976 


9.9986 


26 
^5 


5 


8.8525 


8.8536 


1. 1464 


9.9989 


55 


35 


8.9026 


8.9040 


1.0960 


9.9986 


6 


8.8543 


8.8554 


1. 1446 


9.9989 


54 


36 


8.9042 


8.9056 


1.0944 


9.9986 


24 


7 


8.8560 


8.8572 


1. 1428 


9.9989 


53 


37 


8.9057 


8.9071 


1.0929 


9.9986 


23 


8 


8.8578 


8.8589 


1.1411 


9-9989 


52 


38 


8.9073 


8.9087 


1.0913 


9.9986 


22 


9 


8.8595 


8.8607 


1.1393 


9.9989 


51 


39 


8.9089 


8.9103 


1.0897 


9.9986 


21 


10 


8.8613 


8.8624 


1-1376 


9.9989 


50 


40 


8.9104 


8.9118 


1.0882 


9.9986 


20 


II 


8.8630 


8.8642 


1.1358 


9.9988 


49 


41 


8.9119 


8.9134 


1.0866 


9-9985 


19 


12 


8.8647 


8.8659 


1.1341 


9.9988 


48 


42 


8.9135 


8.9150 


1.0850 


9-9985 


18 


13 


8.8665 


8.8676 


1.1324 


9.9988 


47 


43 


8.9150 


8.9165 


1.0835 


9-9985 


17 


14 


8.8682 


8.8694 


1. 1306 


9.9988 


46 


44 


8.9166 


8.9180 


1.0820 


9-9985 


16 

15 


15 


8.8699 


8.871 1 


1.12*89 


9.9988 


45 


45 


8.9181 


8.9196 


1.0804 


9-9985 


16 


8.8716 


8.8728 


1. 1272 


9.9988 


44 


46 


8.9196 


8.9211 


1.0789 


9-9985 


14 


17 


8.8733 


8.8745 


1.1255 


9.9988 


43 


47 


8.92 1 1 


8.9226 


1.0774 


9-9985 


13 


18 


8.8749 


8.8762 


1.1238 


9.9988 


42 


48 


8.9226 


8.9241 


1-0759 


9-9985 


12 


19 


8.8766 


8.8778 


1. 1222 


9.9988 


41 


49 


8.9241 


8.9256 


1.0744 


9-9985 


II 


20 


8.8783 


8.8795 


1. 1205 


9.9988 


40 


50 


8.9256 


8.9272 


1.0728 


9.9985 


10 


21 


8.8799 


8.8812 


1.1188 


9.9987 


39 


51 


8.9271 


8.9287 


1.0713 


9.9984 


9 


22 


8.8816 


8.8829 


1.1171 


9-9987 


38 


52 


8.9286 


8.9302 


1.0698 


9.9984 


8 


23 


88833 


8.8845 


I-II55 


9.9987 


37 


53 


8.9301 


8.9316 


1.0684 


9.9984 


7 


24 


8.8849 


8.8862 


1.1138 


9.9987 


36 


54 


8.9315 


8:9331 


1.0669 


9.9984 


6 

5 


25 


8.8865 


8.8878 


1.1122 


9.9987 


35 


^^ 


8.9330 


8.9346 


1.0654 


9-9984 


26 


8.8882 


8.8895 


1.1105 


9.9987 


34 


56 


8.9345 


8.9361 


1.0639 


9.9984 


4 


27 


8.8898 


8.8911 


1.1089I 9.9987 


33 


57 


8.9359 


8.9376 


1.0624 


9.9984 


3 


28 


8.8914 


8.8927 


1. 1073 9.9987 


32 


58 


8.9374J 8.9390 


1.0610 


9.9984 


2 


29 


8.8930 


8.8944 


1. 1056 9.9987 


31 


59 


8.9388^ 8.9405 


1.0595 


9-9984 


I 



30 


8.8946 


8.8960 


1.10401 9.9987 


30 


60 


8.9403; 8.9420 


1.0580 


9-9983 




L. Cos. 


L.Cot. L.Tan. L. Sin. 


' 




L.Cos. L.Cot. 


L.Tan. 


L. Sin. 


/ 



85° 



85^ 





L. Sin. 


L.Tan. 


L. Cot. 


L.Cos. 






L. Sin. 


L.Tan. 


L. Cot. 


L.Cos. 




5° 

lO^ 

20^ 

30^ 
40^ 
50^ 

6° 

10^ 
20^ 

30^ 
40^ 
50^ 


8:9403 

8.9545 
8.9682 

8.9816 

8.9945 
9.0070 


8.9420 

8.9563 
8.9701 

8.9836 
8.9966 
9-0093 


1.0580 
1.0437 
1.0299 

1.0164 

1.0034 
0.9907 


9-9983 
9.9982 
9.9981 

9.9980 
9.9979 
9-9977 


85° 

50^ 
40^ 

30^ 
20^ 
10^ 


12° 

10^ 
20'' 

30' 
40' 
50^ 


9-3179 
9-3238 
9-3296 

9-3353 
9.3410 
9-3466 


9-3275 
9-3336 
9-3397 

9-3458 
9-3517 
9-3576 


0.6725 
0.6664 
0.6603 

0.6542 
0.6483 
0.6424 


9-9904 
9.9901 

9-9899 
9.9896 
9-9893 
9.9890 


78° 
50^ 
40^ 

30' 
20' 
10' 


9.0192 
9.031 1 
9.0426 

9-0539 
9.0648 

9-0755 


9.0216 
9-0336 
9-0453 

9-0567 
9.0678 
9.0786 


0.9784 
0.9664 
0.9547 

0.9433 
0.9322 
0.9214 


9.9976 
9-9975 
9-9973 
9.9972 
9.9971 
9.9969 


84° 

50^ 
40^ 

30^ 
20^ 
10^ 


13° 

10^ 
20^ 

30^ 
40^ 
50^ 


9-3521 
9-3575 
9.3629 

9-3682 
9-3734 
9-3786 


9-3634 
9.3691 

9.3748 
9.3804 
9-3859 
9-3914 


0.6366 
0.6309 
0.6252 

0.6196 
0.6141 
0.6086 


9.9887 
9.9884 
9.9881 

9-9878 
9-9875 
9-9872 


77° 
50^ 
40^ 

30/ 
20^ 
10^ 


10^ 
20^ 

30' 
40^ 
50' 


9.0859 
9.0961 
9.1060 

9-I157 
9-1252 
9.1345 


9.0891 

9-0995 
9.1096 

9.1194 
9.1291 
9-1385 


0.9109 
0.9005 
0.8904 

0.8806 
0.8709 
0.8615 


9.9968 
9.9966 
9-9964 

9-9963 
9.9961 

9-9959 


83° 

50^ 
40^ 

30' 
20^ 
10^ 


14° 

10^ 
20^ 

30^ 
40^ 
50^ 


9-3837 
9-3887 
9-3937 
9.3986 

9-4035 
9-4083 


9.3968 
9.4021 
9-4074 
9.4127 
9.4178 
9-4230 


0.6032 

0.5979 
0.5926 

0.5873 
0.5822 
0.5770 


9.9869 
9.9866 
9.9863 

9-9859 
9.9856 

9-9853 


76° 

50^ 
40/ 

30; 
20' 
10^ 


8° 
10' 
20^ 

30' 

40^ 

50^ 


9-1436 
9-1525 
9.1612 

9.1697 
9.1781 
9.1863 


9.1478 

9-1569 
9.1658 

9-1745 
9.1831 

9-1915 


0.8522 9.9958 
0.8431 9.9956 
0.8342 9-9954 
0.8255! 9.9952 
0.8169 9.9950 
0.8085 9-9948 


82° 
50^ 
40^ 

30' 
20^ 
10' 


15° 

10' 
20^ 

30^ 
40^ 

50' 


9.4130 
9.4177 
9-4223 

9.4269 
9.4314 
9-4359 


9.4281 
9-4331 
9-4381 

9.4430 
. 9-4479 
9.4527 


0.5719 
0.5669 
0.5619 

0.5570 
0.5521 
0.5473 


9-9849 
9.9846 
9-9843 

9-9839 
9.9836 
9-9832 


75° 

50^ 

40/ 

30' 
20^ 
10' 


9° 

10^ 
20^ 

30^ 
40^ 
50' 


9.1943 
9.2022 
9.2100 

9.2176 
9.2251 
9.2324 


9.1997 
9.2078 
9.2158 

9.2236 

9.2313 
9.2389 


0.8003 
0.7922 
0.7842 

0.7764 
0.7687 
0.761 1 


9-9946 
9.9944 
9.9942 

9.9940 
9-9938 
9.9936 


81° 
50' 
40^ 

30^ 
20^ 
10' 


16° 

10' 
20^ 

30; 
40' 
50^ 


9.4403 
9.4447 
9.4491 

9-4533 
9-4576 
9.4618 


9.4575 
9.4622 
9.4669 

9.4716 
9.4762 
9.4808 


0.5425 
0.5378 
0.5331 
0.5284 
0.5238 
0.5192 


9.9828 
9-9825 
9.9821 

9.9817 
9.9814 
9.9810 


74° 

50' 
40^ 

30^ 
20^ 
10' 


10° 

10^ 
20' 

30^ 
40^ 
50^ 


9-2397 
9.2468 

9-2538 

9.2606 

9.2674 
9.2740 


9.2463 
9.2536 
9.2609 

9.2680 
9.2750 
9.2819 


0.7537 
0.7464 

0.7391 

0.7320 
0.7250 
0.7181 


9-9934 
9.9931 
9.9929 

9-9927 
9.9924 
9.9922 


80° 

50^ 
40^ 

30^ 
20^ 
10^ 


17° 
10^ 
20^ 

30' 
40^ 

50' 


9-4659 
9.4700 

9-4741 

9-4781 
9.4821 
9.4861 


9-4853 
9.4898 
9-4943 

9-4987 
9-5031 
9-5075 


0.5147 
0.5102 

0.5057 

0.5013 
0.4969 
0.4925 


9.9806 
9.9802 
9-9798 

9-9794 
9-9790 
9-9786 


73° 

50^ 

40^ 

30^ 
20^ 
10^ 


11° 

10^ 
20^ 

30' 
40^ 
50^ 


9.2806 
9:2870 
9.2934 

9-2997 
9-3058 
9-3119 


9.2887 

9-2953 
9.3020 

9-3085 
9-3149 
9.3212 


0.7113 
0.7047 
0.6980 

0.6788 


9.9919 

9-9917 
9.9914 

9.9912 

9-9909 
9.9907 


^90 

50^ 
40' 

30^ 
20^ 
10^ 


18° 
10^ 
20^ 

30^ 
40^ 
50^ 


9.4900 
9-4939 
9-4977 

9-5015 
9-5052 
9.5090 


9.5118 
9.5161 
9-5203 

9-5245 
9-5287 
9-5329 


0.4882 

0.4839 
0.4797 

0.4755 
0.4713 
0.4671 


9.9782 
9-9778 
9.9774 
9.9770 

9-9765 
9.9761 


72° 
50^ 
40' 

30' 
20^ 
10^ 


12° 


9-3179 


9-3275 


0.6725 


9-9904 


78° 


19° 


9.5126 


9-5370 


0.4630 


9-9757 


71° 




I.. Cos. 


L. Cot. 


L.Tan. 


L. Sin. 






L.Cos. 


L. Cot. 


L..Tan 


L. Sin. 







L. Sin. L. Tan. 


L. Cot. 


L.Cos. 






L. Sin. L.Tan. 


L.Cot.jL.Cos. 




19° 

lo' 
20^ 

30/ 
40^ 
50^ 

20° 

10^ 
20^ 

30; 

40^ 

50^ 


9.5126 9.5370 
9.5163 9.541 1 
9-5199 9-5451 

9-5235 9-5491 
9.5270 9.5531 
9-53o6| 9-5571 


0.4630 
0.4589 
0.4549 

0.4509 
0.4469 
0.4429 


9-9757 
9-9752 
9-9748 

9.9743 
9-9739 
9-9734 


n° 

50' 
40^ 

30^ 
20^ 
10^ 


26° 

10^ 
20^ 

30' 
40^ 
50^ 


9.6418 

9-6444 
9.6470 

9-6495 
9.6521 
9-6546 


9.6882 
9.6914 
9.6946 

9.6977 
9.7009 
9.7040 


0.3118 
0.3086 
0.3054 
0.3023 
0.2991 
0.2960 


9-9537 
9.9530 
9-9524 

9.9518 
9.9512 
9.9505 


64° 

50^ 
40/ 

30/ 
20' 
10' 


9-5341 
9-5375 
9-5409 

9-5443 
9-5477 
9-5510 


9.561 1 
9-5650 
9-5689 

9-5727 
9-5766 
9.5804 


0.4389 
0.4350 
0.431 1 

0.4273 

0.4234 
0.4196 


9-9730 
9-9725 
9.9721 

9.9716 
9.971 1 
9.9706 


70° 

50^ 
40/ 

30' 
20^ 
10^ 


2r 
10^ 
20^ 

30/ 
40^ 
50^ 


9-6570 
9-6595 
9.6620 

9.6644 
9.6668 
9.6692 


9.7072 
9.7103 
9-7134 

9-7165 
9.7196 
9.7226 


0.2928 
0.2897 
0.2866 

0.2835 
0.2804 
0.2774 


9.9499 
9.9492 
9.9486 

9.9479 
9-9473 
9.9466 


63° 

50' 
40/ 

30^ 
20^ 
10^ 


21° 

10^ 
20^ 

30; 

40^ 

50^ 


9-5543 
9-5576 
9-5609 

9-5641 
9-5673 
9-5704 


9-5842 
9-5879 
9-5917 

9-5954 
9-5991 
9.6028 


0.4158 
0.4121 
0.4083 

0.4046 
0.4009 
0.3972 


9.9702 

9.9697 
9.9692 

9.9687 
9.9682 
9-9677 


69° 

50' 
40^ 

30' 
20^ 
10^ 


28° 
10^ 
20^ 

30^ 
40^ 

50^ 


9.6716 
9.6740 
9-6763 

9-6787 
9.6810 

9-6833 


9-7257 
9-7287 

9-7317 

9-7348 
9-7378 
9-7408 


0.2743 9-9459 
0.2713 9.9453 
0.2683 9.9446 

0.2652 9.9439 
0.2622 9.9432 
0.2592 9-9425 


62° 

50' 
40^ 

30^ 
20^ 
10^ 


22° 
10^ 
20^ 

30; 
40^ 

50^ 

23° 
10^ 
20^ 

30^ 
40' 
50^ 


9-5736 
9-5767 
9-5798 
9-5828 

9-5859 
9.5889 


9.6064 
9.6100 
9.6136 

9.6172 
9.6208 
9.6243 


0.3936 
0.3900 
0.3864 

0.3828 
0.3792 
0.3757 


9.9672 
9.9667 
9.9661 

9.9656 
9.9651 
9.9646 


68° 
50^ 
40^ 

30^ 
20^ 
10^ 


29° 

10^ 
20^ 

30^ 
40^ 

50^ 


9-6856 
9.6878 
9.6901 

9.6923 
9.6946 
9.6968 


9.7438 0.2562 
9.7467 0.2533 
9.7497 0.2503 

9.7526 0.2474 
9.7556; 0.2444 
9.7585; 0.2415 


9.9418 
9.941 1 
9-9404 

9-9397 
9.9390 

9-9383 


61° 

50^ 
40^ 

30^ 
20^ 
10^ 


9-5919 
9-5948 
9-5978 

9.6007 
9.6036 
9.6065 


9.6279 

9-6314 
9.6348 

9-6383 
9.6417 
9.6452 


0.3721 
0.3686 
0.3652 

0.3617 
0.3583 
0.3548 


9.9640 

9-9635 
9.9629 

9.9624 
9.9618 
9-9613 


6r 
50^ 
40^ 

30^ 
20^ 
10^ 


30° 

10' 

20^ 

30^ 
40^ 

50^ 


9.6990 
9.7012 
9-7033 

9-7055 
9.7076 
9.7097 


9.7614 
9-7644 
9-7673 
9.7701 
9.7730 
9-7759 


0.2386 
0.2356 
0.2327 

0.2299 
0.2270 
0.2241 


9-9375 
9-9368 
9.9361 

9-9353 
9.9346 
9.9338 


60° 

50^ 
40^ 

30^ 
20^ 
10' 


24° 

10^ 
20^ 

30^ 
40^ 
50' 


9.6093 
9.6121 
9.6149 

9.6177 
9.6205 
9.6232 


9.6486 
9.6520 
9-6553 
9.6587 
9.6620 
9.6654 


0.3514 
0.3480 

0.3447 

0.3413 
0.3380 

0.3346 


9-9607 
9.9602 
9-9596 

9-9590 
9-9584 
•9-9579 


66° 

50'" 
40^ 

30^ 
20^ 
10^ 


31° 

10; 
20^ 

30^ 
40^ 
50^ 


9.7118 

9-7139 
9.7160 

9.7181 
9.7201 
9.7222 


9.7788 
9.7816 
9-7845 

9-7873 
9.7902 
9-7930 


0.2212 
0.2184 
0.2155 
0.2127 
0.2098 
0.2070 


9-9331 
9-9323 
9-9315 

9-9308 
9.9300 
9.9292 


59° 

50; 
40^ 

30^ 
20^ 
10^ 


25° 

10' 
20^ 

30/ 
40^ 
50^ 


9-6259 
9.6286 

9-6313 

9-6340 
9.6366 
9.6392 


9.6687 
9.6720 
9-6752 

9-6785 
9.6817 
9-6850 


0.3313 
0.3280 
0.3248 

0.3215 
0.3183 
0.3150 


9-.9573 
9-9567 
9-9561 

9-9555 
9-9549 
9-9543 


65° 

50^ 
40^ 

30^ 
20^ 
10^ 


32° 

10^ 
20^ 

30' 
40^ 

50^ 


9.7242 
9.7262 
9.7282 

9-7302 
9-7322 
9-7342 


9-7958 
9-7986 
9.8014 

9.8042 
9.8070 
9-8097 


0.2042 
0.2014 
0.1986 

0.1958 
0.1930 
0.1903 


9.9284 
9.9276 
9.9268 

9.9260 
9-9252 
9-9244 


58° 
50; 

40^ 

30^ 
20^ 
10^ 


26° 


9.6418 


9.6882 


0.3118 


9-9537 


64° 


33° 


9.7361 


9-8125 


0.1875 


9.9236 


57° 




L. Cos. 


L. Cot. 


L.Tan. 


L. Sin. 






L. Cos. 


L. Cot. 


L.Tan. 


L. Sin. 







L. Sin. L. Tan. 


L. Cot. 


L. Sin. 






L. Sin. 


L.Tan. 


L. Cot. 


L. Cos. 


^^■" 


38° 

20^ 


9-7361 
9.7380 
9.7400 


9-8125 

9-8153 
9.8180 


0.1875 
0.1847 
0.1820 


9.9236 
9.9228 
9.9219 


57° 
50^ 
40^ 


39° 

10^ 
20^ 


9-7989 
9.8004 
9.8020 


9.9084 
9.9110 
9-9135 


0.0916 
0.0890 
0.0865 


9.8905 

9.8895 
9.8884 


51° 

50^ 
40^ 


30^ 
50^ 


9.7419 
9-7438 
9-7457 


9.8208 

9-8235 
9.8263 


0.1792 
0.1765 
0.1737 


9.92 1 1 
9.9203 
9.9194 


30/ 
20^ 
10^ 


30^ 
40^ 
50^ 


9-8035 
9.8050 
9.8066 


9.9161 
9.9187 
9.9212 


0.0839 
0.0813 
0.0788 


9-8874 
9.8864 
9.8853 


30^ 
20^ 
10^ 


34° 

20^ 


9.7476 
9-7494 
9-7513 


9.8290 
9.8317 
9.8344 


0.1710 
0.1683 
0.1656 


9.9186 
9.9177 
9.9169 


56° 

50^ 
40' 


40° 

10^ 
20^ 


9.8081 
9.8096 
9.81 1 1 


9.9238 
9.9264 
9.9289 


0.0762 
0.0736 
0,0711 


9.8843 
9-8832 
9.8821 


50° 

50^ 
40^ 


30^ 
40^ 
50^ 


9-7531 
9-7550 
9.7568 


9.8371 
9.8398 
9.8425 


0.1629 
0.1602 
0.1575 


9.9160 

9-9151 
9.9142 


30/ 
20^ 
10^ 


30/ 
40^ 
50^ 


9-8125 
9.8140 
9-8155 


9-9315 
9-9341 
9.9366 


0.0685 
0.0659 
0.0634 


9.8810 
9.8800 
9.8789 


30; 
20^ 
10^ 


35° 

20^ 


9.7586 
9.7604 
9.7622 


9.8452 

9-8479 
9.8506 


0.1548 
0.1521 
0.1494 


9.9134 
9.9125 
9.9116 


55° 

50^ 
40^ 


41° 

10^ 
20^ 


9.8169 
9.8184 
9.8198 


9-9392 
9.9417 
9-9443 


0.0608 
0.0583 
0.0557 


9-8778 
9-8767 
9-8756 


49° 

50^ 
40/ 


30^ 
40^ 
50' 


9.7640 
9.7657 
9-7675 


9-8533 
9-8559 
9.8586 


0.1467 
0.1441 
0.1414 


9.9107 


30^ 
20^ 
10^ 


30^ 
40^ 
50^ 


9-8213 
9.8227 
9.8241 


9.9468 
9.9494 
9-9519 


0.0532 
0.0506 
0.0481 


9-8745 
9-8733 
9.8722 


30^ 
20^ 
10^ 


36° 

20^ 


9.7692 
9.7710 
9.7727 


9-8613 
9-8639 
9.8666 


0.1387 
0.1361 

0.1334 


9.9080 
9.9070 
9.9061 


54° 

50^ 
40^ 


42° 

10^ 
20' 


9-8255 
9.8269 
9.8283 


9-9544 
9-9570 
9-9595 


0.0456 
0.0430 
0.0405 


9.871 1 


48° 

50^ 
40^ 


30^ 
40^ 
50' 


9.7744 
9.7761 
9-7778 


9.8692 
9.8718 
9.8745 


0.1308 
0.1282 
0.1255 


9.9052 
9.9042 
9.9033 


30^ 
20^ 
10^ 


30^ 
40^ 

50^ 


9.8297 
9.831 1 
9.8324 


9.9621 
9.9646 
9.9671 


0.0379 
0.0354 
0.0329 


9.8676 
9-8665 
9-8653 


30^ 
20^ 
10' 


37° 

20^ 


9-7795 
9.781 1 
9.7828 


9.8771 

9-8797 
9.8824 


0.1229 
0.1203 
0.1 176 


9.9023 
9.9014 
9.9004 


53° 

50^ 
40' 


43° 

10^ 
20^ 


9-8338 
9-8351 
9-8365 


9-9697 
9.9722 

9-9747 


0.0303 
0.0278 
0.0253 


9.8641 
9.8629 
9.8618 


47° 
50^ 
40^ 


30^ 
40^ 
50^ 

38° 
20^ 


9-7844 
9.7861 
9-7877 


9.8850 
9.8876 
9.8902 


0.1150 
0.1124 
0.1098 


9.8995 
9.8985 
9.8975 


30^ 
20^ 
10^ 


30^ 
40/ 
50^ 


9-8378 
9-8391 
9-8405 


9-9772 
9-9798 
9.9823 


0.0228 
0.0202 
0.0177 


9.8606 
9-8594 
9-8582 


30' 
20^ 
10^ 


9-7893 
9.7910 
9.7926 


9.8928 

9-8954 
9.8980 


0.1072 
0.1046 
0.1020 


9.8965 
9-8955 
9.8945 


52° 

50^ 
40^ 


44° 

10^ 
20^ 


9.8418 

9-8431 

9.8444 


9.9848 
9-9874 
9-9899 


0.0152 
0.0126 

O.OIOI 


9-8569 
9-8557 
9-8545 


46° 

50' 

40^ 


30^ 
50^ 


9.7941 
9-7957 
9-7973 


9.9006 
9.9032 
9.9058 


0.0994 
0.0968 
0.0942 


9.8935 
9.8925 
9.8915 


30^ 
20^ 
10^ 


30' 
40^ 

50^ 


9-8457 
9.8469 
9.8482 


9-9924 
9-9949 
9-9975 


0.0076 
0.0051 
0.0025 


9-8532 
9.8520 

9-8507 


30^ 
20^ 
10^ 


39° 


9.7989 


9.9084 


0.0916 


9.8905 


51° 


45° 


9-8495 


0.0000 


0.0000 


9-8495 


45° 




L. Cos. 


L. Cot. 


L.Tan. 


L. Cos. 






L.Cos. 


L. Cot. 


L.Tan. 


L. Sin. 





TABLE III 



NATURAL SINES, COSINES, TANGENTS AND 
COTANGENTS. 



0- 


N. Sin. 


N.Tan. 


N. Cot. 


N. Cos. 






N. Sin. 


N.Tan. 


N. Cot. 


N.Cos. 




.0000 


.0000 


Infin. 


Unity 


90° 


5° 


.0872 


.0875 


11.43 


.9962 


85° 


ic/ 


.0029 


,0029 


343.8 




50^ 


10^ 


.0901 


.0904 


11.06 


.9959 


50^ 


20^ 


.0058 


.0058 


171-9 




40^ 


20^ 


.0929 


.0934 


10.71 


.9957 


40^ 


30^ 


.0087 


.0087 


1 14.6 


" 


30^ 


30^ 


.0958 


.0963 


10.39 


.9954 


30^ 


4c/ 


.0116 


.0116 


85.94 


.9999 


20^ 


40^ 


.0987 


.0992 


10.08 


.9951 


20^ 


5C/ 


.0145 


.0145 


68.75 


.9999 


10^ 


50^ 


.1016 


.1022 


9.788 


.9948 


10' 


1° 


.0175 


.0175 


57.29 


.9998 


89° 


6° 


.1045 


.1051 


9.514 


.9945 


84° 


10^ 


.0204 


.0204 


49.10 


.9998 


50^ 


10^ 


.1074 


.1080 


9.255 


.9942 


50' 


2C/ 


•0233 


.0233 


42.96 


.9997 


40^ 


20^ 


.1103 


.1110 


9.010 


.9939 


40' 


30^ 


.0262 


.0262 


38.19 


.9997 


30^ 


30^ 


.1132 


.1139 


8.777 


.9936 


30^ 


40^ 


.0291 


.0291 


34.37 


.9996 


20^ 


40^ 


.1161 


.1169 


8.556 


.9932 


20^ 


50^ 


.0320 


.0320 


31.24 


.9995 


10^ 


50^ 


.1190 


.1198 


8.345 


.9929 


10^ 


2° 


•0349 


.0349 


28.64 


.9994 


88° 


7° 


.1219 


.1228 


8.144 


.9925 


83° 


10^ 


.0378 


.0378 


26.43 


•9993 


50^ 


10^ 


.1248 


.1257 


7.953 


.9922 


50^ 


20^ 


.0407 


.0407 


24.54 


.9992 


40^ 


20^ 


.1276 


.1287 


7.770 


.9918 


40^ 


3C/ 


.0436 


.0437 


22.90 


.9990 


30^ 


30^ 


.1305 


.1317 


7.596 


.9914 


30^ 


4C/ 


.0465 


.0466 


21.47 


.9989 


20^ 


40^ 


.1334 


.1346 


7.429 


.9911 


20^ 


50^ 


.0494 


.0495 


20.21 


.9988 


10^ 


50^ 


•1363 


.1376 


7.269 


.9907 


10^ 


3° 


•0523 


.0524 


19.08 


.9986 


87° 


8° 


.1392 


.1405 


7.115 


.9903 


82° 


ic/ 


.0552 


•0553 


18.07 


.9985 


50^ 


10^ 


.1421 


.1435 


6.968 


.9899 


50' 


20^ 


.0581 


.0582 


17.17 


.9983 


40/ 


20^ 


.1449 


.1465 


6.827 


.9894 


40^ 


30^ 


.0610 


.0612 


16.35 


.9981 


30^ 


30^ 


.1478 


.1495 


6.691 


.9890 


30^ 


40^ 


.0640 


.0641 


15.60 


.9980 


20^ 


40^ 


.1507 


.1524 


6.561 


.9886 


20^ 


50^ 


.0669 


.0670 


14.92 


.9978 


10' 


50^ 


.1536 


.1554 


6.435 


.9881 


10^ 


4° 


.0698 


.0699 


14.30 


.9976 


88° 


9° 


.1564 


.1584 


6.314 


.9877 


81° 


10^ 


.0727 


.0729 


13.73 


.9974 


50^ 


10^ 


.1593 


.1614 


6.197 


.9872 


50^ 


20^ 


.0756 


.0758 


13.20 


.9971 


40^ 


20^ 


.1622 


.1644 


6.084 


.9868 


40^ 


30^ 


.0785 


.0787 


12.71 


.9969 


30^ 


30^ 


.1650 


.1673 


5.976 


.9863 


30^ 


40^ 


.0814 


.0816 


12.25 


.9967 


20^ 


40^ 


.1679 


.1703 


5.871 


.9858 


20^ 


50^ 


.0843 


.0846 


11.83 


.9964 


10^ 


50^ 


.1708 


.1733 


5.769 


.9853 


10^ 


5° 


.0872 


.0875 


11.43 


.9962 


85° 


10° 


.1736 


.1763 


5.671 


.9848 


80° 




N.Cos. 


N. Cot. 


N.Tan. 


N. Sin. 






N. Cos. 


N. Cot. N.Tan. 


N. Sin. 







N. Sin. 


N.Tan. 


N. Cot. 


N.Cos. 






N. Sin. 


N.Tan. 


N. Cot. 


N.Cos. 




10° 

30; 
50' 


.1736 
.1765 
.1794 

.1822 
.1851 
.1880 


.1763 
.1793 
.1823 

.1853 
.1883 
.1914 


5-671 
5.576 
5.485 

5.396 
5.309 
5.226 


.9848 
.9843 
.9838 

.9833 
.9827 
.9822 


80° 

50^ 
40^ 

30^ 
20^ 
10^ 


17° 
10^ 
20' 

30^ 

40^ 

50^ 


.2924 
.2952 
•2979 

.3007 
•3035 
.3062 


•3057 
.3089 
.3121 

.3153 
.3185 
.3217 


3.271 
3.237 
3.204 

3.172 
3.140 
3.108 


.9563 
•9555 
•9546 

.9537 
.9528 
.9520 


73° 
50' 
40^ 

30' 
20^ 
10^ 


11° 

50^ 


.1908 
•1937 
.1965 

.1994 
.2022 
.2051 


.1944 
.1974 
.2004 

•2035 
.2065 
.2095 


5.145 
5.066 

4.989 

4.915 

4.843 
4.773 


.9816 
.9811 
.9805 

.9799 
.9793 

.9787 


79° 

50^ 
40^ 

30^ 
20^ 
10^ 


18° 
10^ 
20^ 

30^ 
40^ 
50^ 


.3090 
.3118 
.3145 

.3173 
.3201 
.3228 


.3249 
.3281 

.3314 

.3346 
.3378 
.3411 


3.078 

3.047 
3.018 

2.989 
2.960 
2.932 


.9511 
.9502 

.9492 

.9483 
.9474 
.9465 


72° 

50^ 
40^ 

30^ 
20^ 
10^ 


12° 

40^ 
50^ 


.2079 
.2108 
.2136 

.2164 
.2193 
.2221 


.2126 
.2156 
.2186 

.2217 
.2247 
.2278 


4.705 
4.638 

4.574 

4.5 1 1 
4.449 
4.390 


.9781 
.9775 
.9769 

.9763 
.9757 
.9750 


78° 

50^ 
40^ 

30^ 
20' 
10^ 


19° 

10^ 
20^ 

30^ 

40^ 
50^ 


.3256 
.3283 
.3311 

.3338 
.3365 
.3393 


.3443 
.3476 
.3508 

.3541 
.3574 
.3607 


2.904 
2.877 
2.850 

2.824 
2.798 

2.773 


.9455 
.9446 
.9436 

.9426 
.9417 
.9407 


71° 

50' 
40^ 

30/ 
20^ 
10^ 


13° 

30/ 
50^ 


.2250 
.2278 
.2306 

.2334 
.2363 
.2391 


.2309 
.2.339 
.2370 

.2401 
.2432 
,2462 


4.331 
4.275 
4.219 

4.165 
4.113 
4.061 


.9744 
.9737 
.9730 

.9724 
.9717 
.9710 


77° 
50^ 
40^ 

30^ 
20^ 
10^ 


20° 

10^ 
20' 

30^ 
40^ 

50^ 


.3420 
.3448 
.3475 
•3502 
.3529 
.3557 


.3640 
.3673 
.3706 

.3739 
.3772 
.3805 


2.747 
2.723 
2.699 

2.675 
2.651 
2.628 


•9397 
•9387 
•9377 

•9367 
.9356 
.9346 


70° 

50^ 
40^ 

30^ 
20^ 
10^ 

69° 

50^ 
40^ 

30; 
20^ 
10^ 


14° 

30^ 
40^ 
50^ 


.2419 
.2447 
.2476 

.2504 
•2532 
.2560 


.2493 
.2524 
.2555 
.2586 
.2617 
.2648 


4.01 1 
3.962 
3.914 

3.867 
3.821 
3.776 


.9703 

.9681 
.9674 
.9667 


76° 

50^ 
40^ 

30' 
20^ 
10^ 


21° 

10^ 
20^ 

30^ 
40^ 

50^ 


.3584 
.3611 
.3638 

.3665 
.3692 
•3719 


.3839 
.3872 
.3906 

.3939 
.3973 
.4006 


2.605 

2.583 
2.560 

2.539 
2.517 
2.496 


.9336 
.9325 
.9315 

.9304 
.9293 
.9283 


15° 

20^ 

30; 
40^ 

50^ 


.2588 
.2616 
.2644 

.2672 
.2700 
.2728 


.2679 
.2711 
.2742 

.2773 
.2805 
.2836 


3.732 
3.689 
3.647 
3.606 
3.566 
3.526 


.9659 
.9652 
.9644 

.9636 
.9628 
.9621 


75° 
50^ 
40^ 

30' 
20^ 
10^ 


22° 
10' 
20^ 

30' 

40^ 

50^ 


.3746 
.3773 
.3800 

.3827 
.3854 
.3881 


.4040 

.4074 
.4108 

.4142 
.4176 
.4210 


2.475 
2.455 
2.434 

2.414 
2.394 
2.375 


.9272 
.9261 
.9250 

.9239 
.9228 
.9216 


68° 

50' 
40^ 

30^ 
20' 
10^ 


16° 

20' 
30^ 

40^ 
50^ 


.2756 

.2784 
.2812 

.2840 
.2868 
.2896 


.2867 
.2899 
.2931 

.2962 

.2994 
.3026 


3.487 
3.450 
3.412 

3.376 
3.340 
3.305 


.9613 
.9605 
.9596 

.9588 
.9580 
.9572 


74° 
50; 
40^ 

30' 
20' 
10' 


23° 

10^ 
20^ 

30^ 
40^ 
50' 


.3907 
.3934 
.3961 

.3987 
.4014 
.4041 


.4245 
.4279 
.4314 

.4348 
.4383 
.4417 


2.356 
2.337 
2.318 

2.300 
2.282 
2.264 


.9205 

.9194 
.9182 

.9171 
.9159 
.9147 


67° 

50^ 
40^ 

30' 
20^ 
10' 


17° 


.2924 


.3057 


3.271 


•9563 


73° 


24° 


.4067 


•4452 


2.246 


•9135 


66° 




N. Cos. 


N. Cot. 


N.Tan. 


N. Sin. 






N. Cos. 


N. Cot. 


N.Tan. 


N. Sin. 





xxyn 





N. Sin. N. Tan. 

i 


N. Cot. 


N.Cos. 






N. Sin. 'N.Tan. 

1 


N. Cot. N.Cos. 

1 




24° 


.4067 

.4094 
.4120 

.4147 
.4173 
.4200 


.4452 
.4487 
.4522 

.4557 
.4592 
.4628 


2.246 
2.229 
2.211 

2.194 
2.177 
2.161 


.9135 
.9124 
.9112 

.9100 

.9088 
.9075 


66° 

50^ 
40^ 

30^ 
20' 
10^ 


31° 

10^ 
20^ 

30; 
40^ 
50' 


.5150 
•5175 
.5200 

.5225 
.5250 
.5275 


.6009 
.6048 
.6088 

.6128 
.6168 
.6208 


1.664 

1.653 
1.643 

1.632 
1. 62 1 
1.611 


■8572 
•8557 
.8542 

.8526 
.8511 
.8496 


59° 

50^ 
40^ 

30/ 
20^ 
10^ 


25° 

50^ 


.4226 

.4253 
.427t> 

.4305 
.4331 
.4358 


.4663 
.4699 
•4734 

.4770 
.4806 
.4841 


2.145 
2.128 
2.112 

2.097 

2.081 
2.066 


.9063 

.9051 
.9038 

.9026 
.9013 
.9001 


65° 

50^ 
40^ 

30' 
20^ 
10^ 


32° 

10^ 
20^ 

30/ 
40^ 
50^ 


.5299 
•5324 
.5348 

.5373 
•5398 
.5422 


.6249 
.6289 
.6330 

.6371 
.6412 

.6453 


1.600 

1.590 
1.580 

1.570 
1.560 
1.550 


.8480 
.8465 
.8450 

.8434 
.8418 
.8403 


58° 
50^ 
40^ 

30/ 
20^ 
10^ 


26° 

50^ 


.4384 
.4410 
.4436 

.4462 

.4488 
.4514 


.4877 
.4913 
•4950 

.4986 
.5022 
•5059 


2.050 

2.035 
2.020 

2.006 
1. 991 
1.977 


.8988 

.8975 
.8962 

.8949 
.8936 
•8923 


64° 

50^ 
40^ 

30^ 
20^ 
10^ 


33° 

10^ 
20^ 

40^ 
50^ 


•5446 
.5471 
•5495 

.5519 
.5544 
.5568 


.6494 
.6536 
.6577 
.6619 
.6661 
.6703 


1.540 
1.530 
1.520 

1.511 
1.501 
1.492 


•8387 
.8371 
•8355 

.8339 
.8323 
.8307 


57° 
50^ 
40/ 

30^ 
20^ 
10^ 


21° 

40^ 
50^ 


•4540 
.4566 
.4592 
.4617 

.4643 
.4669 


•5095 
•5132 
.5169 
.5206 

.5243 
.5280 


1.963 
1.949 
1-935 
1. 92 1 
1.907 
1.894 


.8910 

.8870 
.8857 
.8843 


63° 

50^ 
40^ 

30; 
20^ 
10^ 


34° 

10^ 
20^ 

30^ 
40^ 
50^ 


.5592 
.5616 
•5640 

.5664 

.5688 
.5712 


.6745 
.6787 
.6830 

.6873 
.6916 

.6959 


1.483 
1.473 
1.464 

1.455 
1.446 
1.437 


.8290 
.8274 
.8258 

.8241 
.8225 
.8208 


56° 
50^ 
40^ 

30^ 
20^ 
10' 

55° 

50^ 
AC/ 

30^ 

2& 


28° 
20^ 

40^ 
50^ 


.4695 
.4720 
.4746 

.4772 
.4797 
.4823 


•5317 
.5354 
•5392 

.5430 
•5467 
.5505 


1.881 
1.868 
1.855 
1.842 
1.829 
1.816 


.8829 
.8816 
.8802 

.8788 

.8774 
.8760 


62° 

50^ 
40^ 

30^ 
20^ 
10' 


35° 

i& 

20^ 

30^ 
40^ 
50^ 


•5736 
.5760 

.5783 

.5807 
.5831 
.5854 


.7002 
.7046 
.7089 

.7133 
.7177 
.7221 


1.428 

1.419 
1.411 

1.402 
1-393 
1.385 


.8192 

•8175 
.8158 

.8141 
.8124 
.8107 


29° 

2C/ 

30; 
40^ 


.4848 

.4874 
.4899 

.4924 
.4950 
.4975 


.5543 
.5581 
.5619 

.5658 
.5696 
•5735 


1.804 
1.792 
1.780 

1.767 
1.756 
1.744 


.8746 
•8732 
.8>i8 

.8704 
.8689 
•8675 


61° 

50^ 
40' 

30; 
20' 
10^ 


36° 

10^ 
20^ 

30^ 
40^ 
50^ 


.5878 
•5901 
.5925 

•5948 
•5972 
•5995 


.7265 
.7310 
•7355 
.7400 
•7445 
.7490 


1.376 
1.368 
1.360 

1.351 
1.343 
1.335 


.8090 

•8073 
.8056 

.8039 
.8021 
.8004 


54° 

50^ 
40^ 

30; 
20^ 
10^ 


30° 

2(y 

30' 
40; 


.5000 
.5025 
•5050 

.5075 
.5100 
.5125 


•5774 
.5812 

.5851 
.5890 
•5930 
•5969 


1.732 
1.720 
1.709 

1.698 
1.686 
1.675 


.8660 
.8646 
.8631 

.8616 
.8601 
.8587 


60° 

50^ 
40' 

30^ 
20^ 
10' 


37° 

10^ 
20^ 

30' 
40^ 
5C/ 


.6018 
.6041 
.6065 

.6088 
.6111 
.6134 


•7536 
.7581 
.7627 

.7673 
.7720 
.7766 


1.327 

1-319 
1.311 

1-303 
1-295 

1.288 


.7986 
•7969 
.7951 

•7934 
.7916 
.7898 


53° 

50^ 
40^ 

30^ 
20^ 
10^ 

52° 


31° 


•5150 


.6009 


1.664 


.8572 


59° 


38° 


.6157 


•7813 


1.280 .7880 




N.Cos. 


N. Cot. 


N.Tan. 


N. Sin. 






N.Cos. 


N.Cot.^N.Tan.N.Sin. 

1 1 







N. Sin. 


N.Tan. 


N. Cot. 


N. Cos. 




38° 
50^ 


•6157 
.6180 
.6202 

.6225 
.6248 
.6271 


.7813 
.7860 

.7907 

•7954 
.8002 
.8050 


1.280 
1.272 
1.265 

1.257 
1.250 
1.242 


.7880 
.7862 
.7844 
.7826 
.7808 
.7790 


52° 

50^ 
40^ 

30^ 
20^ 
10^ 


39° 

50' 


.6293 
.6316 
.6338 
.6361 

.6406 


.8098 
.8146 
.8195 
.8243 
.8292 
.8342 


1-235 
1.228 
1.220 

1.213 
1.206 
1.199 


.7771 
•7753 
.7735 
.7716 
.7698 
.7679 


51° 

50^ 
40^ 

30^ 
20^ 
10^ 


40° 

20^ 

30; 
40^ 
50' 


.6428 
.6450 
.6472 

.6494 
•6517 
•6539 


•8391 
.8441 
.8491 

.8541 

.8591 
.8642 


1. 192 
1.185 
1. 178 

1. 171 
1. 164 
1.157 


.7660 
.7642 
.7623 

.7604 

•7585 
.7566 


50° 

50^ 
40^ 

30; 
20^ 
10^ 


41° 

20^ 

30^ 
40' 
50^ 


.6561 

.6583 
.6604 

.6626 
.6648 
.6670 


.8693 

.8744 
.8796 

.8847 
.8899 
.8952 


1. 150 
1. 144 
I-I37 
1. 130 
1. 124 
1.117 


.7547 
.7528 

.7509 

.7490 
.7470 
.7451 


49° 

50^ 
40^ 

30^ 
20^ 
10^ 


42° 

20^ 

30' 
40^ 
50^ 


.6691 
.6713 
.6734 

.6756 
.6777 
.6799 


.9004 

.9057 
.9110 

■9163 
.9217 
.9271 


I, III 
1. 104 
1.098 

1.091 
1.085 
1.079 


.7431 
.7412 

.7392 

.7373 
.7353 
•7333 


48° 

50^ 
40^ 

30' 
20^ 
10^ 


43° 

20^ 

30^ 

40^ 
50^ 


.6820 
.6841 
.6862 

.6884 
.6905 
.6926 


•9325 
.9380 

•9435 

.9490 
.9545 
.9601 


1.072 
1.066 
1.060 

1.054 
1.048 
1.042 


•7314 
•7294 

•7274 

.7254 
•7234 
.7214 


47° 

50^ 
40^ 

30^ 
20^ 
10^ 


44° 

20' 

30' 

40^ 
50^ 


•6947 .9657 
.6967! .9713 
.6988 .9770 

.7009 .9827 
.7030 .9884 
.7050 .9942 


1.036 
1.030 
1.024 

1.018 
1. 01 2 
1.006 


•7193 
•7173 
•7153 

.7133 
.7112 
.7092 


46° 

50^ 
40^ 

30^ 
20^ 
10' 


45° 


.7071 1. 0000 


1. 000 


.7071 


45° 




N. Cos. N. Cot. 


N.Tan. 


N. Sin. 





OCT 10 1902 



OCT 10 1902 



